What is the integral of 8/(x^2sqrt(x^2+16)) ?

The integral of 8/(x^2sqrt(x^2+16)) is -(sqrt(16+x^2))/(2x)+C

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integral of 8/(x^2sqrt(x^2+16))

\int \frac{8}{x ^{2} x ^{2} + 16} d x

- \frac{16 + x ^{2}}{2 x} + C

Solution steps

\int \frac{8}{x ^{2} x ^{2} + 16} 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} + 16} d x

Apply Trigonometric Substitution

: \int \frac{sec ( u )}{16 tan ^{2} ( u )} d u

= 8 \cdot \int \frac{sec ( u )}{16 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}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 8 \cdot \frac{1}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 8 \cdot \frac{1}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 8 \cdot \frac{1}{16} (- \frac{1}{v})

Substitute back

= 8 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})

Simplify

8 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}) : - \frac{16 + x ^{2}}{2 x}

= - \frac{16 + x ^{2}}{2 x}

Add a constant to the solution

= - \frac{16 + x ^{2}}{2 x} + C

x \to \infty lim (81 x^{2} + x - 9 x)

\int 7 e^{9 x} d x

\int y \cdot e^{- y} d y

(2 y + 2 t - 9) d t + (8 y + 2 t - 1) d y = 0

\frac{d}{d x} (\frac{7 x}{x - 1})

What is the integral of 8/(x^2sqrt(x^2+16)) ?
The integral of 8/(x^2sqrt(x^2+16)) is -(sqrt(16+x^2))/(2x)+C