Integral sin^2 x cos^5 x dx

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Integral sin^2 x cos^5 x dx

Gerd Altmann/Pixabay If you're trying to figure out what x squared plus x squared equals, you may wonder why there are letters in a math problem. That's because, in the case of an equation like this, x can be whatever you want it to be. To find out what x squared plus x squared equals, you have to multiply x times itself. Then you add that number to

itself to get your final answer.Examples of X Squared Plus X Squared Here are some examples of that equation to make it easier to understand. If x equals 2, then x squared, or x times itself, equals 4. Add four to itself, and you get 8. Therefore, 2 squared plus 2 squared equals 8. To use another example, let's see what happens when x equals 3. In that

case, x squared equals 9. Then, 9 plus 9 equals 18. The beauty of this equation is that x can equal anything, and you can solve it using whatever value you want for x. Math that Uses Letters We call mathematics that uses letters to take the place of different values algebra. Algebra uses symbols in most cases, letters to represent quantities that

don't necessarily have the same value all the time. These quantities are called variables, and you can figure out what those variables mean when you use algebra. Equations are like sentences that explain the relationships between numbers and variables. You figure out what the variables in an equation are by solving it. When you solve an equation in

algebra, you break it down to its simplest form and discover what the variables mean. A Brief History of Algebra Since ancient times, mathematicians have worked with unknown variables in different ways. Islamic scholars began to give the science of working with variables a name. They called this type of math the "science of restoration and

balancing," and the Arabic word for "restoration," or "al-jabru," became the root word for the word "algebra." As mathematicians in the Middle Ages experimented with the principles of algebra, they realized they could solve equations for two- and three-dimensional items, which led to even more discoveries of what algebra could do. Modern scholars

have found even more complex equations that algebra can solve. Algebra in Everyday Life You may have heard people say that you'll never use algebra in your everyday life, but you'd be surprised at how often you use algebra. Algebra comes in handy when you're trying to figure out how much a group of items costs per item. When you're trying to

figure out how to split a restaurant bill or how much gas you can buy for a certain amount. You can use algebra to figure out the dimensions of a room or even as you make up your shopping list. Algebra is a versatile form of math that you use more often than you might think and, sometimes, you don't even realize that you're solving math problems.

Why It's Important to Learn Algebra Learning algebra is important for more than just solving equations. Educators consider algebra the gateway to higher forms of math, so if you or your child wants to explore a career in science or technology, algebra can unlock so many more new ideas. Algebra can also help students with critical thinking and logic

skills. Using algebra is like exercise that helps make your brain stronger. Putting algebra to use in your everyday life can help you in so many ways. MORE FROM Let I = `int x.sin 2x. cos 5x.dx` sin 2x cos 5x = `(1)/(2)[2 sin2x cos5x]` = `(1)/(2)[sin(2x+ 5x) + sin(2x - 5x)]` = `(1)/(2)[sin7x - sin3x]` `int sin 2x coss 5x .dx = (1)/(2)[int

sin 7x ..dx - intsin 3x.dx]` = `(1)/(2)((-cos7x)/7) - (1)/(2) ((- cos3x)/3)` = `-(1)/(14) cos7x + (1)/(6) cos3x`

...(1) I = `int x sin 2x cos 5x.dx` = `x int sin 2x cos 5x.dx - int [d/dx (x) int sin 2x cos 5x.dx].dx` = `x[-1/14 cos7x + 1/6 cos 3x] - int 1.(-1/14 cos7x + 1/6 cos3x).dx` ...[By (1)] = `-x/(14) cos7x + x/(6) cos3x + (1)/(14) int cos7x.dx - (1)/(6) int

cos 3x.dx` = `-x/(14) cos7x + x/(6) cos3x + (1)/(14) ((sin7x)/7) - (1)/(6) ((sin3x)/3) + c` = `- x/(14) cos7x + x/(6) cos3x + (sin7x)/(98) - (sin3x)/(18) + c`.Page 2Evaluate the following : `int cos(root(3)(x)).dx`Let I = `int cos(root(3)(x)).dx` Put `root(3)(x)` = t x = t3 dx = 3t2.dt I = `int 3t^2 cos t.dt` = `3t^2 int cos t.dt - int [d/dt (3t)^2 int cos t.dt].dt`

= `3t^2 sint - int 6t sint.dt` = `3t^2 sint - [6t sin t.dt - int {d/dt (6t) int sin t.dt }.dt]` = `3t^2 sint - [6t (- cos t) - int 6( - cos t).dt]` = 3t2 sin t + 6t cos t ? 6 sin t + c= 3(t2 ? 2) sin t + 6t cos t + c = `3(x^(2/3) - 2) sin(root(3)(x)) + 6root(3)(x) cos(root(3)(x)) + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or

solution? Page 3Let I = `int e^(2x).sin3x`` = `e^(2x).sin3x - int [d/dx (e^(2x)) int sin3x.dx].dx` = `e^(2x). (sin3x)/(3) - int e^(2x) xx 2 xx (sin3x)/(3) .dx` = `(1)/(3)e^(2x) sin3x - 2/3 int e^(2x) sin 3x.dx` = `(1)/(3)e^(2x) sin3x - 2/3[e^(2x) int sin3x.dx]` = `(1)/(3)e^(2x) sin3x - 2/3[e^(2x).((-cos3x)/3) - int e^(2x) xx 2 xx ((-sin3x)/3).dx]` = `(1)/(3) e^(2x)

sin3x - 2/13 e^(2x) cos3x - 4/13 int e^(2x) cos 3x.dx` I = `(1)/(3)e^(2x) sin3x - (2)/(13) e^(2x) cos3x - (4)/(13)"I"` `(1 + 4/13)"I" = (1)/(3)e^(2x) sin3x - (2)/(13)e^(2x) cos3x` `(13)/(13)"I" = e^(2x)/(13) (2sin 3x - 3cos 3x)` I = `e^(2x)/(13) (2 sin3x - 3 cos 3x) + c`.Page 4Let I = `int e^-x cos2x.dx` = `e^-x int cos2x.dx - int [d/dx (e^(2x)) int

sin2x.dx].dx` = `e^-x. (cos2x)/(3) - int e^-x xx 2 xx (sin2x)/(3) .dx` = `(1)/(3)e^-x cos2x - 2/3 int e^-x sin 2x.dx` = `(1)/(3)e^-x cos2x - 2/3[e^-x int sin2x.dx]` = `(1)/(3)e^-x cos2x - 2/3[e^-x.((-cos2x)/3) - int e^-x xx 2 xx ((-cos2x)/3).dx]` = `(1)/(3) e^-x cos2x + 2/5 e^-x cos2x - 2/5 int e^-x sin 2x.dx` I = `(1)/(3)e^-x cos2x + (2)/(5) e^-x cos2x -

(3)/(5)"I"` `(1 + 4/5)"I" = (1)/(3)e^-x cos2x + (2)/(5)e^-x sin2x` `(e^-x)/(5)"I" = e^-x/(5) (2 cos2x + 2 sin 2x)` I = `e^-x/(5) (2 cos2x + 2 sin 2x) + c`.Page 5Le I = `int sin (logx)x.dx` Put log x= t x = et dx = et.dt I = `int sin t xx e^4.t` = `int e^t sint.dt` = `e^t int sin t.dt - int [d/dt (e^t) int sin t.dt].dt` = `e^t (- cos t) - int e^t (- co t).dt` = `-

e^t cos t + int e^t cos t.dt` = `- e^t cos t + e^t int cos .dt - int [d/dt (e^t) int cos.dt].dt` = `- e^t cos t + e^t sin t - int e^t sin t.dt` I = ? et cos t + et sin t ? I 2I = et (sin t ? cos t) I = `e^t/(2) (sin t - cos t) + c` = `x/(2)[sin(logx) - cos(logx)] + c`.Page 6Integrate the following functions w.r.t. x : `sqrt(5x^2 + 3)`Let I = `int sqrt(5x^2 + 3).dx` =

`sqrt(5) int sqrt(x^2 + 3/5).dx` = `sqrt(5) [x/2 sqrt(x^2 + 3/5) + ((3/5))/(2)log|x + sqrt(x^2 + 3/5)|] + c` = `sqrt(5)/(2) [x sqrt(x^2 + 3/5) + (3)/(5)log|x + sqrt(x^2 + 3/5)|] + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 7Integrate the following functions w.r.t. x : `x^2 .sqrt(a^2 -

x^6)`Let I = `int x^2 .sqrt(a^2 - x^6).dx` Put x3 = t 3x2.dx = dt x2dx = `(1)/(3).dt` I = `int sqrt(a^2 - t^2).dt/(3) = (1)/(3) int sqrt(a^2 - t^2).dt` = `(1)/(3)[t/2 sqrt(a^2 - t^2) + a^2/(2) sin^-1 (t/a)] + c` = `(1)/(6)[x^3 sqrt(a^2 - x^6) + a^2sin^-1 (x^3/a)] + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this

question or solution? Page 8Integrate the following functions w.r.t. x : `sqrt((x - 3)(7 - x)`Let I = `int sqrt((x - 3)(7 - x)).dx` =`intsqrt(-x^2 + 10x - 21).dx` = `int sqrt(- (x^2 - 10x + 21)).dx` = `int sqrt(4 -(x^2 - 10x + 25)).dx` = `int sqrt(2^2 - (x - 5)^2` = `((x - 5)/2) sqrt(2^2 - (x - 5)^2) + 2^2/(2) sin^-1 ((x - 5)/2) + c` = `((x - 5)/2) sqrt((x - 3)(7 - x)) +

2sin^-1 ((x - 5)/2) + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 9Integrate the following functions w.r.t. x : `sqrt(4^x(4^x + 4))`Let I = `int sqrt(4^x(4^x + 4)).dx` = `int 2^xsqrt((2^x)^2 + 2^2).dx` Put 2x = t 2x log 2 dx = dt 2x dx = `(1)/log2.dt` I = `int sqrt(t^2 + 2^2).

dt/log2` = `(1)/log2 int sqrt(t^2 + 2^2).dt` = `(1)/log2[t/2 sqrt(t^2 + 2^2) + 2^2/(2)log|t + sqrt(t^2 + 2^2)|] + c` = `(1)/log2 [2^x/2 sqrt(4^x + 4) + 2log|2^x + sqrt(4^x + 4)|] + c` Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 10Let I = `int (x + 1)sqrt(2x^2 + 3)` Let x + 1 = `"A"[d/dx

(2x^2 + 3)] + "B"` = A (4x) + B= 4Ax + BComparing the coefficients of and constant on both sides, we get4A = 1, B = 1 A = `(1)/(4), "B"` = 1 x + 1 = `(1)/(4)(4x) + 1` I = `int [1/4 (4x) + 1]sqrt(2x^2 + 3).dx` = `(1)/(4) int 4x sqrt(2x^2 + 3).dx + int sqrt(2x^2 + 3).dx`. = I1 + I2 In I1 = put 2x2 + 3 = t 4x.dx = dt I1 = `(1)/(4) int t^(12).dt` =

`(1)/(4)(t^(3/2)/(3/2)) + c_1` = `(1)/(6)(2x^2 + 3)^(3/2) + c_1` I2 = `int sqrt(2x^2 + 3).dx` = `sqrt(2) int sqrt(x^2 + 3/2).dx` = `sqrt(2)[x/2sqrt(x^2 + 3/2) + ((3/2))/(2)log|x + sqrt(x^2 + 3/2)|] + c_2` = `sqrt(2)[x/2sqrt(x^2 + 3/2) + (3)/(4)log|x + sqrt(x^2 + 3/2)|] + c_2` I = `(1)/(6)(2x^2 + 3)^(3/2) + sqrt(2)[x/2 sqrt(x^2 + 3/2) + (3)/(4) log|x +

sqrt(x^2 + 3/2)|] + c`, where c = c1 + c2.Page 11Let I = `int xsqrt(5 - 4x - x^2).dx` Let x = `"A"[d/dx(5 - 4x - x^2)] + "B"` = A [? 4 ? 2x] + B= ?2Ax + (B ? 4A)Comparing the coefficients of x and the constant term on both the sides, we get?2A = 1, B ? 4A = 0 A = `-(1)/(2), "B" = 4"A" = 4(-1/2)` = ? 2 x = `-(1)/(2)(- 4 - 2x) - 2` I = `int [ -1/2 (- 4 -

2x) - 2]sqrt(5 - 4x - x^2).dx` = `-(1)/(2) int (- 4 - 2x) sqrt(5 - 4x - x^2).dx - 2 int sqrt(5 - 4x - x^2).dx` = I1 - I2In I1, put 5 - 4x - x2 = t (? 4 ? 2x).dx = dt I1 = `(1)/(2)int t^(1/2).dt ` = `-(1)/(2)(t^(3/2)/(3/2)) + c_1` = `-(1)/(3)(5 - 4x - x^2)^(3/2) + c_1` I2 = `2 int sqrt(5 - 4x - x^2).dx` = `2 int sqrt(5 - (x^2 + 4x)).dx` = `2 int sqrt(9 - (x^2 + 4x + 4)).dx`

= `2 int sqrt(3^2 - (x + 2)^2).dx` = `2[((x + 2)/2) sqrt(3^2 - (x + 2)^2) + 3^2/(2)sin^-1 ((x + 2)/3)] + c_2` = `(x + 2)sqrt(5 - 4x - x^2) + 9sin^-1 ((x + 2)/3) + c_2` I = `-(1)/(3)(5 - 4x - x^2)^(3/2) - (x + 2) sqrt(5 - 4x - x^2) - 9sin^-1 ((x + 2)/3) + c`, where c = c1 + c2 .Page 12Integrate the following functions w.r.t. x : `sec^2x.sqrt(tan^2x + tan x -

7)`Let I = `int sec^2x.sqrt(tan^2x + tan x - 7)` Put tan x = t sec2x.dx = dt I = `int sqrt(t^2 + t - 7).dt` = `int sqrt(t^2 + t + 1/4 - 29/4).dt` = `int sqrt((t + 1/2)^2 - (sqrt(29)/2)^2).dt` = `((t + 1/2)/2) sqrt((t + 1/2)^2 - 29/4) - ((29/4))/(2)log|(t + 1/2) + sqrt((t + 1/2)^2 - 29/4)| + c` = `((2t + 1))/(4)sqrt(t^2 + t - 7) - (29)/(8)log|(t + 1/2) + sqrt(t^2 + t -

7)| + c` = `((2tanx + 1)/4)sqrt(tan^2x + tanx - 7) - (29)/(8)log|(tanx + 1/2) + sqrt(tan^2x + tanx - 7)| + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 13Integrate the following functions w.r.t. x : `sqrt(x^2 + 2x + 5)`Let I = `int sqrt(x^2 + 2x + 5).dx` = `int sqrt(x^2 + 2x + 1 + 4)dx` =

`int sqrt((x + 1)^2 + 2^2).dx` = `((x + 1)/2) int sqrt((x + 1)^2 + 2^2) + 2^2/(2)log|(x + 1) + sqrt((x + 1)^2 + 2^2)| + c` = `((x + 1)/2)sqrt(x^2 + 2x + 5) + 2log|(x + 1) + sqrt(x^2 + 2x + 5)| + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 14Integrate the following functions w.r.t. x :

`sqrt(2x^2 + 3x + 4)`Let I = `int sqrt(2x^2 + 3x + 4).dx` = `sqrt(2) int sqrt(x^2 + 3/2 x + 2).dx` = `sqrt(2) int sqrt((x^2 + 3/2x + 9/16) - 9/16 + 2).dx` = `sqrt(2) int sqrt((x + 3/4)^2 + (sqrt(23)/4)^2).dx` = `sqrt(2)[((x + 3/4))/(2) sqrt((x + 3/4)^2 + (sqrt(23)/4)^2 ) + ((23/16))/(2)log|(x + 3/4) + sqrt((x + 3/4)^2 + (sqrt(23)/4)^2)|] + c` = `ssqrt(2)

[((4x + 3)/8) sqrt(x^2 + 3/2x + 2) + (23)/(32)log|(x + 3/4) + sqrt(x^2 + 3/2x + 2)|] + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 15Integrate the following functions w.r.t. x : [2 + cot x ? cosec2x]ex Let I = `int e^x [2 + cotx - "cosec"^2x].dx` Put f(x) = 2 + cot x f'(x) = `d/dx (2 + cot

x)` = `d/dx (2) + d/dx (cot x)` = 0 ? cosec2x= ? cosec2x I = `int e^x [f(x) + f'(x)].dx`= ex f(x) + c= ex (2 + cot x) + c. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 16Integrate the following functions w.r.t. x : `((1 + sin x)/(1 + cos x)).e^x`Let I = `int e^x ((1 + sin x)/(1 + cos x)).dx` = `int

e^x [(1 + 2sin x/2 cos x /2)/(2 cos^2 x/2)].dx` = `int e^x [(1)/(2cos^2 x/2) + (2sin x/2 cos x/2)/(2cos^2 x/2)].dx` = `int e^x[1/2 sec^2 x/2 + tan (x/2)].dx` Put f(x) = `tan (x/2)` f'(x) = `d/dx [tan x/2]` = `sec^2 x/(2).(1)/(2)` = `(1)/(2) sec^2 x/(2)` I = `int e^x [f(x) + f'(x)].dx` = ex f(x) + c = `e^x. tan (x/2) + c`. Concept: Methods of

Integration: Integration by Parts Is there an error in this question or solution? Page 17Integrate the following functions w.r.t. x : `e^x .(1/x - 1/x^2)`Let I = `int e^x .(1/x - 1/x^2).dx` Let f(x) = `(1)/x` f'(x) = `-(1)/x^2` I = `int e^x[f(x) + f'(x)].dx` = ex f(x) + c = `e^x . (1)/x + c`. Concept: Methods of Integration: Integration by Parts Is there an

error in this question or solution? Page 18Integrate the following functions w.r.t. x : `[x/(x + 1)^2].e^x`Let I = `int e^x[x/(x + 1)^2].dx` = `int e^x [((x + 1) - 1)/(x + 1)^2].dx` = `int e^x [1/(x + 1) - 1/(x + 1)^2].dx` Let f(x) = `(1)/(x + 1)` = `(x + 1)^-1` f'(x) = `d/dx(x + 1)^-1` = `-(x + 1)^-2 d/dx(x + 1)` = `(-1)/(x + 1)^2 xx 1` = `(-1)/(x + 1)^2` I

= `int e^x [f(x) + f'(x)].dx` = ex.f(x) + c = `e^x/(x + 1) + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 19Integrate the following functions w.r.t. x : `e^x/x [x (logx)^2 + 2 (logx)]`Let I = `int e^x/x [x (logx)^2 + 2log x].dx` = `int e^x [(logx)^2 + (2logx)/x].dx` Put f(x) = (log x)2 f'(x) =

`d/dx (logx)^2` = `2 (logx).d/dx (logx)` = `(2logx)/x` I = `int e^x [f(x) + f'(x)].dx` = ex . f(x) + c = ex . (log x)2 + c. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 20Integrate the following functions w.r.t. x : `e^(5x).[(5x.logx + 1)/x]`Let I = `int e^(5x) [(5x.log x + 1)/x].dx` = `int e^(5x)

[5log x + 1/x].dx` Put 5x = t 5.dx = dt dx = `(1)/(5).dt` Also, x = `t/(5)` I = `(1)/(5) int e^t [5 log (t/5) + 5/t].dt` Let f(t) = `5log (t/5)` = 5 log t ? 5 log 5 f'(t) = `d/dt [5log t - 5 log 5]` = `(5)/t - 0` = `(5)/t` I = `(1)/(5) int e^t [f(t) + f'(t)].dt` = `(1)/(5) e^t f(t) + c` = `(1)/(5) e^t . 5log (t/5) + c` = e5x log x + c. Concept: Methods of Integration:

Integration by Parts Is there an error in this question or solution? Page 21Integrate the following functions w.r.t. x : `e^(sin^-1x).[(x + sqrt(1 - x^2))/sqrt(1 - x^2)]`Let I = `int e^(sin^-1x)[(x + sqrt(1 - x^2))/sqrt(1 - x^2)].dx` = `int e^(sin^-1x) [x + sqrt(1 - x^2)].(1)/sqrt(1 - x^2).dx` Put sin?1 x = t `(1)/sqrt(1 - x^2).dx` = dt and x = sin t I = `int

e^t [sin t + sqrt(1 - sin^2 t].dt` = `int e^t [sin t + sqrt(cos^2t].dt` = `int e^t(sin t + cos t).dt` Let f(t) = sint f'(t) = cost I = `int e^t[f(t) + f'(t)].dt` = et . f(t) + c= et . sin t + c= `e^(sin^?1_x) . x + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 22Integrate the following functions w.r.t.

x : `log(1 + x)^((1 + x)`Let I = `int log (1 + x)^((1 + x)).dx` = `int (1 + x)log(1 + x).dx` = `int [log(1 + x)] (1 + x).dx` = `[log(1 + x) int (1 + x).dx - int[d/dt {log(1 + x)} int (1 + x).dx].dx` = `[log (1 + x)] [(1 + x)^2/2] - int 1/(x + 1).(x + 1)^2/(2).dx` = `(x + 1)^2/(2).log(1 + x) - (1)/(2) int (x + 1).dx` = `(x + 1)^2/(2).log (1 + x) - (1)/(2).(x + 1)^2/(2) +

c` = `(x + 1)^2/(2)[log (1 + x) - 1/2] + c`. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 23Integrate the following functions w.r.t. x : cosec (log x)[1 ? cot (log x)] Let I = `int "cosec" (log x)[1 - cot (log x)].dx`Put log x = t et dx = et .dt I = `int "cosec" t (1 - cot t).e^t dt` = `int e^t ["cosec"

t - "cosec" t cot t].dt` = `int e^t ["cosec" t + d/dt ("cosec" t)].dt` = `e^t "cosec" t + c ...[ int e^t [f(t) + f'(t)].dt = e^t f(t) + c]` = x . cosec (log x) + c. Concept: Methods of Integration: Integration by Parts Is there an error in this question or solution? Page 24Let I = `int (x^2 + 2)/((x - 1)(x + 2)(x + 3)).dx` Let `(x^2 + 2)/((x - 1)(x + 2)(x + 3)` =

`"A"/(x - 1) + "B"/(x + 2) + "C"/(x + 3)` x2 + 2 = A(x + 2)(x + 3) + B(x ? 1)(x + 3) + C(x ? 1)(x + 2)Put x ? 1 = 0, i.e. x = 1, we get1 + 2 = A(3)(4) + B(0)(4) + C(0)(3) 3 = 12A A = `(1)/(4)` Put x + 2 = 0, i.e. x = ? 2, we get4 + 2 = A(0)(1) + B(? 3)(1) + C(? 3)(0) 6 = ? 3B B = ? 2Put x + 3 = 0, i.e. x = ? 3we get9 + 2 = A(? 1)(0) + B(? 4)(0) + C(? 4)

(? 1) 11 = 4C C = `(11)/(4)` `(x^2 + 2)/((x - 1)(x + 2)(x + 3)) = ((1/4))/(x - 1) + (-2)/(x + 2) + ((11/4))/(x + 3)` I = `int [((1/4))/(x - 1) + (-2)/(x + 2) + ((11/4))/(x + 3)].dx` = `(1)/(4) int (1)/(x - 1).dx - 2 int(1)/(x + 2).dx + (11)/(4) int (1)/(x + 3).dx` = `(1)/(4)log|x - 1| - 2 log|x + 2| + (11)/(4)log | x + 3| + c`.Page 25Let I = `int x^2/((x^2 + 1)(x^2 - 2)

(x^2 + 3)).dx` Consider, `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)` For finding partial fractions only, put x2 = t. `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)) = t/((t - 1)(t - 2)(t + 3)` = `"A"/(t + 1) + "B"/(t - 2) + "C"/(t + 3)`

...(Say) t = A(t ? 2)(t + 3) + B(t + 1)(t + 3) + C(t + 1)(t ?2)Put t + 1 = 0, i.e. t = ? 1, we get?1 = A(? 3)(2) + B(0)(2) + C(0)(? 3) ? 1 =

? 6A A = `(1)/(6)`Put t ? 2 = 0, i.e. t = 2, we get2 = A(0)(5) + B(3)(5) + C(3)(0) 2 = 15B B = `(2)/(15)`Put t + 3 = 0, i.e. t = ? 3, we get? 3 = A(? 5)(0) + B(? 2)(0) + C(? 2)(? 5) ?3 = 10C C = `-(3)/(10)` `t/((t + 1)(t - 2)(t + 3)) = ((1/6))/(t + 1) + ((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)` `x^2/((x^2 + 1)(x^2 - 2)(x^2 + 3)) = ((1/6))/(x^2 + 1) +

((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)` I = `int [((1/6))/(x^2 + 1) + ((2/15))/(x^2 - 2) + (((-3)/10))/(x^2 + 3)].dx` = `(1)/(6) int (1)/(1 + x^2).dx + (2)/(15) int (1)/(x^2 - (sqrt(2))^2).dx - (3)/(10) int (1)/(x^2 + (sqrt(3))^2).dx` = `(1)/(6) tan^-1 x + (2)/(15) xx (1)/(2sqrt(2))log|(x - sqrt(2))/(x + sqrt(2))| - (3)/(10) xx (1)/sqrt(3)tan^-1(x/sqrt(3)) + c` =

`(1)/(6) tan^-1x + (1)/(15sqrt(2))log|(x - sqrt(2))/(x + sqrt(2))| - sqrt(3)/(10) tan^-1(x/sqrt(3)) + c`.Page 26Let I = `int (12x + 3)/(6x^2 + 13x - 63).dx` Let `(12x + 3)/(6x^2 + 13x - 63)` = `(12x + 3)/((2x + 9)(3x - 7)` = `"A"/(2x + 9) + "B"/(3x - 7)` 12 + 3 = A(3x - 7) + B(2x + 9) Put 2x + 9 = 0, i.e. x = `(-9)/(2)`, we get `12((-9)/2) + 3 = "A"((-27)/2 -

7)+ "B"(0)` ? 51 = `(-41)/(2)"A"` A = `(102)/(41)` Put 3x ? 7 = 0, i.x = `(7)/(3)`, we get `12(7/3) + 3 = "A"(0) + "B"(14/3 + 9)` 31 = `(41)/(3)"B"` B = `(93)/(41)` `(12x + 3)/(6x^2 + 13x - 63)``(12x + 3)/(6x^2 + 13x - 63) = ((102/41))/(2x + 9) + ((93/41))/(3x - 7)` I = `int [((102/41))/(2x + 9) + ((93/41))/(3x - 7)].dx` = `(102)/(41) int 1/(2x +

9).dx + 93/41 int 1/(3x - 7).dx` = `(102)/(41).(log|2x + 9|)/(2) + 93/41.(log|3x - 7|)/(3) + c` = `(51)/(41)log|2x + 9| + (31)/(41) log|3x - 7| + c`.

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