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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

= \frac{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

= \frac{1}{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

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

Add a constant to the solution

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

\frac{d}{d x} (y + x (ln (x) - ln (y) - 1))

\int 4 x - 5 d x

\frac{s ^{3}}{s + 2}

\frac{d}{d x} (ln (2 x - y))

\int_{0}^{1} 5 d x

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