Question

If \(I_n = \int \frac{1}{(x^2+1)^n} dx\), then \(2n I_{n+1} - (2n-1) I_n =\)

• A. \(\frac{(x^2+1)^n}{x} + c\)
• B. \(\frac{x}{(x^2+1)^n} + c\)
• C. \(x(x^2+1)^{n-1} + c\)
• D. \(\frac{x}{(x^2+1)^{n-1}} + c\)

Hint:

Reduction formulas for integrals of the form \(\frac{1}{(x^2+a^2)^n}\) are standard. Remembering the technique of adding and subtracting constants in the numerator during integration by parts is crucial.

Answer 1

The Correct Option is B

Step 1: Use Integration by Parts on \(I_n\):
Let \(u = \frac{1}{(x^2+1)^n}\) and \(dv = dx\). Then \(du = -n(x^2+1)^{-n-1}(2x) dx\) and \(v = x\). \[ I_n = \frac{x}{(x^2+1)^n} - \int x \left( \frac{-2nx}{(x^2+1)^{n+1}} \right) dx \] \[ I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{x^2}{(x^2+1)^{n+1}} dx \]
Step 2: Algebraic Manipulation:
Write \(x^2\) in the numerator as \((x^2+1) - 1\): \[ I_n = \frac{x}{(x^2+1)^n} + 2n \int \frac{(x^2+1) - 1}{(x^2+1)^{n+1}} dx \] \[ I_n = \frac{x}{(x^2+1)^n} + 2n \left[ \int \frac{1}{(x^2+1)^n} dx - \int \frac{1}{(x^2+1)^{n+1}} dx \right] \]
Step 3: Relate \(I_n\) and \(I_{n+1}\):
Substitute definitions of \(I_n\) and \(I_{n+1}\): \[ I_n = \frac{x}{(x^2+1)^n} + 2n [ I_n - I_{n+1} ] \] \[ I_n = \frac{x}{(x^2+1)^n} + 2n I_n - 2n I_{n+1} \] Rearrange terms to match the required expression: \[ 2n I_{n+1} = \frac{x}{(x^2+1)^n} + 2n I_n - I_n \] \[ 2n I_{n+1} = \frac{x}{(x^2+1)^n} + (2n - 1) I_n \] \[ 2n I_{n+1} - (2n - 1) I_n = \frac{x}{(x^2+1)^n} \] Final Answer:
The value is \(\frac{x}{(x^2+1)^n} + c\).