What is the integral of (4x^2)/(x^2+8) ?

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

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integral of (4x^2)/(x^2+8)

\int \frac{4 x ^{2}}{x ^{2} + 8} d x

- 82 arctan (\frac{x}{2 2}) + 4 x + C

Solution steps

\int \frac{4 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

= 4 \cdot \int \frac{x ^{2}}{x ^{2} + 8} d x

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

= 4 \cdot \int - \frac{8}{x ^{2} + 8} + 1 d x

Apply Integral Substitution

= 4 \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

= 4 \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

= 4 \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

= 4 \cdot 22 (- arctan (u) + u)

Substitute back

u = \frac{x}{2 2}

= 4 \cdot 22 (- arctan (\frac{x}{2 2}) + \frac{x}{2 2})

Simplify

4 \cdot 22 (- arctan (\frac{x}{2 2}) + \frac{x}{2 2}) : - 82 arctan (\frac{x}{2 2}) + 4 x

= - 82 arctan (\frac{x}{2 2}) + 4 x

Add a constant to the solution

= - 82 arctan (\frac{x}{2 2}) + 4 x + C

y = (\frac{( x - 1 )}{( x - 2 )})^{2}

\frac{d}{d x} (- csc (x) cot (x))

\frac{d y}{d x} + \frac{e}{π} y = 5

\frac{d y}{d x} + y^{3} x + 7 y = 0

y^{^{''}} - 3 y^{^{'}} + 2 y = 2 t^{2} + e^{t} + 1

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