5A. Inverse trigonometric functions; Hyperbolic

E. Solutions to 18.01 Exercises 5. Integration techniques 5A-3 2a) y = x − 1, so 1 − y2 = 4x/(x + 1) , and 1 = (x + 1) . Hence x + 1 1 − y2 2 √ x dy 2 = dx (x + 1)2 d dy/dx ................
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