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

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

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

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

- \frac{1}{2} arctan (x) - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{x - 1} + C

Solution steps

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

Expand

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

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

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

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

Simplify

- \frac{1}{2 ( x - 1 )} - \frac{1}{2} arctan (x) - \frac{1}{2} ln ∣ x - 1 ∣ - \frac{1}{2 ( x - 1 )} + \frac{1}{4} ln x^{2} + 1 : - \frac{1}{2} arctan (x) - \frac{1}{2} ln ∣ x - 1 ∣ + \frac{1}{4} ln x^{2} + 1 - \frac{1}{x - 1}

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

Add a constant to the solution

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

\int x^{2} a^{2} + x^{2} d x

\int \frac{1}{1 - 16 y ^{2}} d y

\int e^{2 y} (4 - e^{2 y})^{\frac{4}{3}} d y

\int (e^{y^{2}} y) d y

\int \frac{- 3 x ^{2}}{16 - x ^{2}} d x

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