Evaluate:.

OR

Evaluate:

Let I =

_{We can easily observe that the derivative of sin}^-1x is available in expression, so we need to apply the method of substitution to solve this

_{Putting, Sin⁻¹x = t}

⇒

∴ I =

Applying Integrating by parts we have -

I = t (-cos t) –

⇒ I =-t cos t + sin t + c

Putting back the value of t –

⇒ I = . sin⁻¹x + x + C

OR

Let I =

_{By observing the integral, we can clearly see that it has a form that suits the application of Integration by a partial fraction-}

∴

⇒ x² + 1 = A (x – 1) (x + 3) + B (x + 3) + C (x – 1)²

⇒ x² + 1 = A(x² + 2x – 3) + B(x + 3) + C (x² - 2x + 1)

⇒ x² + 1 = x²(A + C) + x(2A + B – 2C) + (C + 3B – 3A)

By equating, we have -

A + C = 1

2A + B – 2C = 0

-3A + 3B + C = 1

∴ A = , B = , C =

∴

⇒ I =

∵

∴ I = log (x – 1) – + log (x+3) + C