What is the integral of (1-x^2)/((x+1)(x-3)^2(2x+1)) ?

The integral of (1-x^2)/((x+1)(x-3)^2(2x+1)) is -3/49 ln|x-3|+2/(7(x-3))+3/49 ln|2x+1|+C

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integral of (1-x^2)/((x+1)(x-3)^2(2x+1))

\int \frac{1 - x ^{2}}{( x + 1 ) ( x - 3 ) ^{2} ( 2 x + 1 )} d x

- \frac{3}{49} ln ∣ x - 3 ∣ + \frac{2}{7 ( x - 3 )} + \frac{3}{49} ln ∣ 2 x + 1 ∣ + C

Solution steps

\int \frac{1 - x ^{2}}{( x + 1 ) ( x - 3 ) ^{2} ( 2 x + 1 )} d x

Take the partial fraction of

\frac{1 - x ^{2}}{( x + 1 ) ( x - 3 ) ^{2} ( 2 x + 1 )} : - \frac{3}{49 ( x - 3 )} - \frac{2}{7 ( x - 3 ) ^{2}} + \frac{6}{49 ( 2 x + 1 )}

= \int - \frac{3}{49 ( x - 3 )} - \frac{2}{7 ( x - 3 ) ^{2}} + \frac{6}{49 ( 2 x + 1 )} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int \frac{3}{49 ( x - 3 )} d x - \int \frac{2}{7 ( x - 3 ) ^{2}} d x + \int \frac{6}{49 ( 2 x + 1 )} d x

\int \frac{3}{49 ( x - 3 )} d x = \frac{3}{49} ln ∣ x - 3 ∣

\int \frac{2}{7 ( x - 3 ) ^{2}} d x = - \frac{2}{7 ( x - 3 )}

\int \frac{6}{49 ( 2 x + 1 )} d x = \frac{3}{49} ln ∣ 2 x + 1 ∣

= - \frac{3}{49} ln ∣ x - 3 ∣ - (- \frac{2}{7 ( x - 3 )}) + \frac{3}{49} ln ∣ 2 x + 1 ∣

Simplify

= - \frac{3}{49} ln ∣ x - 3 ∣ + \frac{2}{7 ( x - 3 )} + \frac{3}{49} ln ∣ 2 x + 1 ∣

Add a constant to the solution

= - \frac{3}{49} ln ∣ x - 3 ∣ + \frac{2}{7 ( x - 3 )} + \frac{3}{49} ln ∣ 2 x + 1 ∣ + C

What is the integral of (1-x^2)/((x+1)(x-3)^2(2x+1)) ?
The integral of (1-x^2)/((x+1)(x-3)^2(2x+1)) is -3/49 ln|x-3|+2/(7(x-3))+3/49 ln|2x+1|+C