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

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

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

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

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

Solution steps

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

Simplify

x^{2} \frac{1}{1 + x ^{2}} : \frac{x ^{2}}{1 + x ^{2}}

= \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

\int (2 x^{2} - x + 3) d x

\int \frac{x ^{2} - x + 5}{x + 3} d x

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

\int (3 x^{2} + 4 x - 5) d x

\int \frac{2 x - 5}{x ^{2} - 5 x + 3} d x

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