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

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

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

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

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

Solution steps

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

Apply Integral Substitution

= \int \frac{2 u ^{2}}{u ^{2} + 1} d u

Take the constant out:

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

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

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

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

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}{u ^{2} + 1} d u + \int 1 d u)

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

\int 1 d u = u

= 2 (- arctan (u) + u)

Substitute back

u = \frac{x}{2}

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

Simplify

2 (- arctan (\frac{x}{2}) + \frac{x}{2}) : - 2 arctan (\frac{x}{2}) + x

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

Add a constant to the solution

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

y = 3 - 6 x

\frac{\partial}{\partial x} (x^{2} cos (x))

y^{^{'}} + x y = 1

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

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

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