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

The integral of (x+1)/((x-1)^3) is -1/(x-1)-1/((x-1)^2)+C

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

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

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

Solution steps

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

Expand

\frac{x + 1}{( x - 1 ) ^{3}} : \frac{x}{( x - 1 ) ^{3}} + \frac{1}{( x - 1 ) ^{3}}

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

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

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

= - \frac{1}{x - 1} - \frac{1}{2 ( x - 1 ) ^{2}} - \frac{1}{2 ( x - 1 ) ^{2}}

Simplify

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

= - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}}

Add a constant to the solution

= - \frac{1}{x - 1} - \frac{1}{( x - 1 ) ^{2}} + C

\int \frac{tan ( x )}{1} d x

y^{^{'}} = t - 2 y

\int_{2}^{\infty} \frac{1}{x ^{4}} d x

f (x) = x^{2} - 7 x + 12, at x = 2

y = 4 - 4 + x

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