What is the integral of (5x^2+x-34)/((x+1)^2(x-4)) ?

The integral of (5x^2+x-34)/((x+1)^2(x-4)) is 3ln|x+1|-6/(x+1)+2ln|x-4|+C

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integral of (5x^2+x-34)/((x+1)^2(x-4))

\int \frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} d x

3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣ + C

Solution steps

\int \frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} d x

Take the partial fraction of

\frac{5 x ^{2} + x - 34}{( x + 1 ) ^{2} ( x - 4 )} : \frac{3}{x + 1} + \frac{6}{( x + 1 ) ^{2}} + \frac{2}{x - 4}

= \int \frac{3}{x + 1} + \frac{6}{( x + 1 ) ^{2}} + \frac{2}{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{3}{x + 1} d x + \int \frac{6}{( x + 1 ) ^{2}} d x + \int \frac{2}{x - 4} d x

\int \frac{3}{x + 1} d x = 3 ln ∣ x + 1 ∣

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

\int \frac{2}{x - 4} d x = 2 ln ∣ x - 4 ∣

= 3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣

Add a constant to the solution

= 3 ln ∣ x + 1 ∣ - \frac{6}{x + 1} + 2 ln ∣ x - 4 ∣ + C

\frac{x}{2}

\frac{d}{d x} (\frac{2 x + 7}{2 x})

- (- x^{3} + y^{3}) d x + x y^{2} d y = 0

\int \frac{1}{3 x ^{2} + 4 x - 7} d x

\frac{d}{d x} (x (x^{5} + 9)^{3})

What is the integral of (5x^2+x-34)/((x+1)^2(x-4)) ?
The integral of (5x^2+x-34)/((x+1)^2(x-4)) is 3ln|x+1|-6/(x+1)+2ln|x-4|+C