What is the integral of (7x^2+13x)/(x^3+x^2-x-1) ?

The integral of (7x^2+13x)/(x^3+x^2-x-1) is 5ln|x-1|+2ln|x+1|-3/(x+1)+C

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integral of (7x^2+13x)/(x^3+x^2-x-1)

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

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

Solution steps

\int \frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} d x

Expand

\frac{7 x ^{2} + 13 x}{x ^{3} + x ^{2} - x - 1} : \frac{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} + \frac{13 x}{x ^{3} + x ^{2} - x - 1}

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{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} d x + \int \frac{13 x}{x ^{3} + x ^{2} - x - 1} d x

\int \frac{7 x ^{2}}{x ^{3} + x ^{2} - x - 1} d x = 7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

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

= 7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 13 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣)

Simplify

7 (\frac{3}{4} ln ∣ x + 1 ∣ + \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) + 13 (- \frac{1}{4} ln ∣ x + 1 ∣ - \frac{1}{2 ( x + 1 )} + \frac{1}{4} ln ∣ x - 1 ∣) : 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1}

= 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1}

Add a constant to the solution

= 5 ln ∣ x - 1 ∣ + 2 ln ∣ x + 1 ∣ - \frac{3}{x + 1} + C

\frac{\partial}{\partial x} (6 x^{\frac{1}{3}} y^{\frac{2}{3}})

\frac{d}{d a} (\frac{cos ^{2} ( a )}{sin ( a ) - 1})

t an g e n t f (x) = x^{4} + 4 x^{2} - x, (1, 4)

\int \frac{x ^{2}}{x ^{2} + 16} d x

n = 1 \sum \infty \frac{1000}{5 ^{n}}

What is the integral of (7x^2+13x)/(x^3+x^2-x-1) ?
The integral of (7x^2+13x)/(x^3+x^2-x-1) is 5ln|x-1|+2ln|x+1|-3/(x+1)+C