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

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

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

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

- \frac{2 x ^{2} + 4}{x} + C

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

= 8 \cdot \int \frac{sec ( u )}{4 tan ^{2} ( u )} d u

Take the constant out:

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

= 8 \cdot \frac{1}{4} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 8 \cdot \frac{1}{4} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

: \int \frac{1}{v ^{2}} d v

= 8 \cdot \frac{1}{4} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{4} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )})

Simplify

8 \cdot \frac{1}{4} (- \frac{1}{sin ( arctan ( \frac{1}{2} x ) )}) : - \frac{2 x ^{2} + 4}{x}

= - \frac{2 x ^{2} + 4}{x}

Add a constant to the solution

= - \frac{2 x ^{2} + 4}{x} + C

\frac{d}{d x} (\frac{e ^{y}}{x + y})

\int (6 x^{2} cos (2 x^{3} + 1)) d x

\frac{d}{d x} (2 sin^{2} (x) + 5 sin (x) - 3)

\frac{d}{d x} (\frac{x}{e ^{2 x}})

y^{^{'}} + 8 x^{2} y = x^{2}

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