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

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

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

integral

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

2 (- arctan (x) + x) + C

Solution steps

\int \frac{2 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

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

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

= 2 \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

= 2 (- \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

= 2 (- arctan (x) + x)

Add a constant to the solution

= 2 (- arctan (x) + x) + C

\frac{1}{cos ^{2} ( 1 - 2 x )}

\frac{1}{2 x ^{2} y}

x^{4} e^{x}

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

e^{(8 x)^{2}}

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