What is the integral of (2x^2-1)/((x+1)^2(x+4)) ?

The integral of (2x^2-1)/((x+1)^2(x+4)) is -13/9 ln|x+1|-1/(3(x+1))+31/9 ln|x+4|+C

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integral of (2x^2-1)/((x+1)^2(x+4))

\int \frac{2 x ^{2} - 1}{( x + 1 ) ^{2} ( x + 4 )} d x

- \frac{13}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{31}{9} ln ∣ x + 4 ∣ + C

Solution steps

\int \frac{2 x ^{2} - 1}{( x + 1 ) ^{2} ( x + 4 )} d x

Take the partial fraction of

\frac{2 x ^{2} - 1}{( x + 1 ) ^{2} ( x + 4 )} : - \frac{13}{9 ( x + 1 )} + \frac{1}{3 ( x + 1 ) ^{2}} + \frac{31}{9 ( x + 4 )}

= \int - \frac{13}{9 ( x + 1 )} + \frac{1}{3 ( x + 1 ) ^{2}} + \frac{31}{9 ( x + 4 )} 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{13}{9 ( x + 1 )} d x + \int \frac{1}{3 ( x + 1 ) ^{2}} d x + \int \frac{31}{9 ( x + 4 )} d x

\int \frac{13}{9 ( x + 1 )} d x = \frac{13}{9} ln ∣ x + 1 ∣

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

\int \frac{31}{9 ( x + 4 )} d x = \frac{31}{9} ln ∣ x + 4 ∣

= - \frac{13}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{31}{9} ln ∣ x + 4 ∣

Add a constant to the solution

= - \frac{13}{9} ln ∣ x + 1 ∣ - \frac{1}{3 ( x + 1 )} + \frac{31}{9} ln ∣ x + 4 ∣ + C

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What is the integral of (2x^2-1)/((x+1)^2(x+4)) ?
The integral of (2x^2-1)/((x+1)^2(x+4)) is -13/9 ln|x+1|-1/(3(x+1))+31/9 ln|x+4|+C