What is the integral of 1/((9x^2+16)^{3/2)} ?

The integral of 1/((9x^2+16)^{3/2)} is x/(16(9x^2+16)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/((9x^2+16)^{3/2)}

\int \frac{1}{( 9 x ^{2} + 16 ) ^{\frac{3}{2}}} d x

\frac{x}{16 ( 9 x ^{2} + 16 ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{1}{( 9 x ^{2} + 16 ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

= \int \frac{1}{48 sec ( u )} d u

Take the constant out:

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

= \frac{1}{48} \cdot \int \frac{1}{sec ( u )} d u

Rewrite using trig identities

= \frac{1}{48} \cdot \int cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= \frac{1}{48} sin (u)

Substitute back

u = arctan (\frac{3}{4} x)

= \frac{1}{48} sin (arctan (\frac{3}{4} x))

Simplify

\frac{1}{48} sin (arctan (\frac{3}{4} x)) : \frac{x}{16 ( 9 x ^{2} + 16 ) ^{\frac{1}{2}}}

= \frac{x}{16 ( 9 x ^{2} + 16 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{x}{16 ( 9 x ^{2} + 16 ) ^{\frac{1}{2}}} + C

What is the integral of 1/((9x^2+16)^{3/2)} ?
The integral of 1/((9x^2+16)^{3/2)} is x/(16(9x^2+16)^{1/2)}+C