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

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

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

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

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

Solution steps

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

Take the partial fraction of

\frac{2 x + 2}{( x ^{2} + 1 ) ( x - 1 ) ^{3}} : \frac{1}{x ^{2} + 1} - \frac{1}{( x - 1 ) ^{2}} + \frac{2}{( x - 1 ) ^{3}}

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

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

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

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

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

Simplify

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

Add a constant to the solution

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

y = (arctan (2 x))^{2}

\frac{d}{d x} (arctan (x^{3}))

\frac{d}{d x} (arcsin (5 x^{2}))

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

y^{^{'}} = (2 y + 1) cot (x)

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