What is the integral of (4x^2)/(1+x^2) ?

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

integral of (4x^2)/(1+x^2)

\int \frac{4 x ^{2}}{1 + x ^{2}} d x

4 (- arctan (x) + x) + C

Solution steps

\int \frac{4 x ^{2}}{1 + x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 4 \cdot \int \frac{x ^{2}}{1 + x ^{2}} d x

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

= 4 \cdot \int - \frac{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

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

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

\int 1 d x = x

= 4 (- arctan (x) + x)

Add a constant to the solution

= 4 (- arctan (x) + x) + C

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