What is the integral of 9/(x^2sqrt(x^2+16)) ?

The integral of 9/(x^2sqrt(x^2+16)) is -(9sqrt(16+x^2))/(16x)+C

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integral of 9/(x^2sqrt(x^2+16))

\int \frac{9}{x ^{2} x ^{2} + 16} d x

- \frac{9 16 + x ^{2}}{16 x} + C

Solution steps

\int \frac{9}{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

= 9 \cdot \int \frac{1}{x ^{2} x ^{2} + 16} d x

Apply Trigonometric Substitution

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

= 9 \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

= 9 \cdot \frac{1}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= 9 \cdot \frac{1}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= 9 \cdot \frac{1}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= 9 \cdot \frac{1}{16} (- \frac{1}{v})

Substitute back

= 9 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )})

Simplify

9 \cdot \frac{1}{16} (- \frac{1}{sin ( arctan ( \frac{1}{4} x ) )}) : - \frac{9 16 + x ^{2}}{16 x}

= - \frac{9 16 + x ^{2}}{16 x}

Add a constant to the solution

= - \frac{9 16 + x ^{2}}{16 x} + C

tan^{2} (5 θ)

n = 1 \sum \infty 5 (- 1)^{n} n x^{n}

e^{- 4}

\frac{d}{d x} ((sin (x^{2}) + 1)^{\frac{1}{f ( x )}})

\int x^{2} x^{2} + 1 d x

What is the integral of 9/(x^2sqrt(x^2+16)) ?
The integral of 9/(x^2sqrt(x^2+16)) is -(9sqrt(16+x^2))/(16x)+C