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

The integral of (2x^2-7x-1)/((x-1)(x+1)^2) is -3/2 ln|x-1|+7/2 ln|x+1|+4/(x+1)+C

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

\int \frac{2 x ^{2} - 7 x - 1}{( x - 1 ) ( x + 1 ) ^{2}} d x

- \frac{3}{2} ln ∣ x - 1 ∣ + \frac{7}{2} ln ∣ x + 1 ∣ + \frac{4}{x + 1} + C

Solution steps

\int \frac{2 x ^{2} - 7 x - 1}{( x - 1 ) ( x + 1 ) ^{2}} d x

Expand

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

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{2 x ^{2}}{( x - 1 ) ( x + 1 ) ^{2}} d x - \int \frac{7 x}{( x - 1 ) ( x + 1 ) ^{2}} d x - \int \frac{1}{( x - 1 ) ( x + 1 ) ^{2}} d x

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

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

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

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

Simplify

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

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

Add a constant to the solution

= - \frac{3}{2} ln ∣ x - 1 ∣ + \frac{7}{2} ln ∣ x + 1 ∣ + \frac{4}{x + 1} + C

y^{^{'}} = \frac{5 x ^{3} + y ^{3}}{x y ^{2}}

\int 3 cos (7 x) d x

y^{^{''}} + 49 y^{^{'}} = 0

\frac{\partial}{\partial a} (cos (a) sin (a + 2 b))

\frac{\partial}{\partial x} (3 x^{2} - 2 x y + y^{2} - 8 y)

What is the integral of (2x^2-7x-1)/((x-1)(x+1)^2) ?
The integral of (2x^2-7x-1)/((x-1)(x+1)^2) is -3/2 ln|x-1|+7/2 ln|x+1|+4/(x+1)+C