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

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

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

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

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

Solution steps

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

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

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

Apply Integral Substitution

= \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

= 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

= 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

= 22 (- arctan (u) + u)

Substitute back

u = \frac{x}{2 2}

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

Simplify

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

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

Add a constant to the solution

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

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

y^{^{''}} - 4 y^{^{'}} + 4 = 0

n = 0 \sum \infty \frac{( 3 ^{n} )}{n !}

\frac{\partial}{\partial x} (x y^{2} - e^{x y})

y^{^{''}} + 20 y^{^{'}} + 200 y = 300, y (0) = 0, y^{^{'}} (0) = 0

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