What is the integral of 6/(x^2sqrt(x^2+16)) ?

The integral of 6/(x^2sqrt(x^2+16)) is -(3sqrt(x^2+16))/(8x)+C

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integral of 6/(x^2sqrt(x^2+16))

\int \frac{6}{x ^{2} x ^{2} + 16} d x

- \frac{3 x ^{2} + 16}{8 x} + C

Solution steps

\int \frac{6}{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

= 6 \cdot \int \frac{1}{x ^{2} x ^{2} + 16} d x

Apply Trigonometric Substitution

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

= 6 \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

= 6 \cdot \frac{1}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 6 \cdot \frac{1}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 6 \cdot \frac{1}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 6 \cdot \frac{1}{16} (- \frac{1}{v})

Substitute back

= 6 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})

Simplify

6 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}) : - \frac{3 x ^{2} + 16}{8 x}

= - \frac{3 x ^{2} + 16}{8 x}

Add a constant to the solution

= - \frac{3 x ^{2} + 16}{8 x} + C

x \to 2 lim (x + 1)

\int_{0}^{\infty} sin (2 x) d x

x \to 2 lim (\frac{3 x + 1}{x + 2})

\frac{d}{d t} (0.00225 t e^{- 0.0467 t})

\frac{\partial}{\partial x} (x y (sin ((x^{5}) + y^{5})))

What is the integral of 6/(x^2sqrt(x^2+16)) ?
The integral of 6/(x^2sqrt(x^2+16)) is -(3sqrt(x^2+16))/(8x)+C