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

The integral of ((3x^2-1))/((x^2+1)) is 3x-4arctan(x)+C

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

integral

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

3 x - 4 arctan (x) + C

Solution steps

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

Expand

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

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

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

= 3 (- arctan (x) + x) - arctan (x)

Simplify

3 (- arctan (x) + x) - arctan (x) : 3 x - 4 arctan (x)

= 3 x - 4 arctan (x)

Add a constant to the solution

= 3 x - 4 arctan (x) + C

x^{2} x + 2

\frac{1 + 1}{x ^{4}}

\frac{e ^{x} + 1}{e ^{x}}

e^{7} y

e^{x^{2} + y^{2}}

What is the integral of ((3x^2-1))/((x^2+1)) ?
The integral of ((3x^2-1))/((x^2+1)) is 3x-4arctan(x)+C