What is the integral of (3x^2+1)/((x+1)(x-5)^2) ?

The integral of (3x^2+1)/((x+1)(x-5)^2) is 1/9 ln|x+1|+26/9 ln|x-5|-(38)/(3(x-5))+C

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integral of (3x^2+1)/((x+1)(x-5)^2)

\int \frac{3 x ^{2} + 1}{( x + 1 ) ( x - 5 ) ^{2}} d x

\frac{1}{9} ln ∣ x + 1 ∣ + \frac{26}{9} ln ∣ x - 5 ∣ - \frac{38}{3 ( x - 5 )} + C

Solution steps

\int \frac{3 x ^{2} + 1}{( x + 1 ) ( x - 5 ) ^{2}} d x

Take the partial fraction of

\frac{3 x ^{2} + 1}{( x + 1 ) ( x - 5 ) ^{2}} : \frac{1}{9 ( x + 1 )} + \frac{26}{9 ( x - 5 )} + \frac{38}{3 ( x - 5 ) ^{2}}

= \int \frac{1}{9 ( x + 1 )} + \frac{26}{9 ( x - 5 )} + \frac{38}{3 ( x - 5 ) ^{2}} 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{1}{9 ( x + 1 )} d x + \int \frac{26}{9 ( x - 5 )} d x + \int \frac{38}{3 ( x - 5 ) ^{2}} d x

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

\int \frac{26}{9 ( x - 5 )} d x = \frac{26}{9} ln ∣ x - 5 ∣

\int \frac{38}{3 ( x - 5 ) ^{2}} d x = - \frac{38}{3 ( x - 5 )}

= \frac{1}{9} ln ∣ x + 1 ∣ + \frac{26}{9} ln ∣ x - 5 ∣ - \frac{38}{3 ( x - 5 )}

Add a constant to the solution

= \frac{1}{9} ln ∣ x + 1 ∣ + \frac{26}{9} ln ∣ x - 5 ∣ - \frac{38}{3 ( x - 5 )} + C

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What is the integral of (3x^2+1)/((x+1)(x-5)^2) ?
The integral of (3x^2+1)/((x+1)(x-5)^2) is 1/9 ln|x+1|+26/9 ln|x-5|-(38)/(3(x-5))+C