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

The integral of (4x^2-1)/(1+x^2) is 4x-5arctan(x)+C

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

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

4 x - 5 arctan (x) + C

Solution steps

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

Expand

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

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

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

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

Simplify

4 (- arctan (x) + x) - arctan (x) : 4 x - 5 arctan (x)

= 4 x - 5 arctan (x)

Add a constant to the solution

= 4 x - 5 arctan (x) + C

What is the integral of (4x^2-1)/(1+x^2) ?
The integral of (4x^2-1)/(1+x^2) is 4x-5arctan(x)+C