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

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

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

\int \frac{x + x ^{2}}{1 + x ^{2}} d x

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

Solution steps

\int \frac{x + x ^{2}}{1 + x ^{2}} d x

\frac{x + x ^{2}}{1 + x ^{2}} = \frac{x - 1}{1 + x ^{2}} + 1

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

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

\int 1 d x = x

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

Add a constant to the solution

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

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\int (x + 8)^{4} d x

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