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

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

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

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

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

Solution steps

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

Expand

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

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

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

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

Add a constant to the solution

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

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