What is the integral of 1/(x^2sqrt(16+9x^2)) ?

The integral of 1/(x^2sqrt(16+9x^2)) is -(sqrt(16+9x^2))/(16x)+C

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integral of 1/(x^2sqrt(16+9x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{3 sec ( u )}{16 tan ^{2} ( u )} d u

= \int \frac{3 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

= \frac{3}{16} \cdot \int \frac{sec ( u )}{tan ^{2} ( u )} d u

Rewrite using trig identities

= \frac{3}{16} \cdot \int \frac{cos ( u )}{sin ^{2} ( u )} d u

Apply u-substitution

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

= \frac{3}{16} \cdot \int \frac{1}{v ^{2}} d v

Apply the Power Rule

: - \frac{1}{v}

= \frac{3}{16} (- \frac{1}{v})

Substitute back

= \frac{3}{16} (- \frac{1}{sin ( arctan ( \frac{3}{4} x ) )})

Simplify

\frac{3}{16} (- \frac{1}{sin ( arctan ( \frac{3}{4} x ) )}) : - \frac{16 + 9 x ^{2}}{16 x}

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

Add a constant to the solution

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

y^{^{''}} - 6 y = 9 t

\int 5 x^{2} + 9 d x

t v^{^{'}} (1 - t^{2}) + 2 v = 0

\int x (x^{2} - 1)^{2} d x

(1 + x)^{n}

What is the integral of 1/(x^2sqrt(16+9x^2)) ?
The integral of 1/(x^2sqrt(16+9x^2)) is -(sqrt(16+9x^2))/(16x)+C