What is the integral of (2x^2)/(x^2+8) ?

The integral of (2x^2)/(x^2+8) is -4sqrt(2)arctan(x/(2sqrt(2)))+2x+C

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integral of (2x^2)/(x^2+8)

\int \frac{2 x ^{2}}{x ^{2} + 8} d x

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

Solution steps

\int \frac{2 x ^{2}}{x ^{2} + 8} 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}}{x ^{2} + 8} d x

\frac{x ^{2}}{x ^{2} + 8} = - \frac{8}{x ^{2} + 8} + 1

= 2 \cdot \int - \frac{8}{x ^{2} + 8} + 1 d x

Apply Integral Substitution

= 2 \cdot \int 22 (- \frac{1}{u ^{2} + 1} + 1) d u

Take the constant out:

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

= 2 \cdot 22 \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 \cdot 22 (- \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 \cdot 22 (- arctan (u) + u)

Substitute back

u = \frac{x}{2 2}

= 2 \cdot 22 (- arctan (\frac{x}{2 2}) + \frac{x}{2 2})

Simplify

2 \cdot 22 (- arctan (\frac{x}{2 2}) + \frac{x}{2 2}) : - 42 arctan (\frac{x}{2 2}) + 2 x

= - 42 arctan (\frac{x}{2 2}) + 2 x

Add a constant to the solution

= - 42 arctan (\frac{x}{2 2}) + 2 x + C

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What is the integral of (2x^2)/(x^2+8) ?
The integral of (2x^2)/(x^2+8) is -4sqrt(2)arctan(x/(2sqrt(2)))+2x+C