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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{3 cos ^{2} ( u )} d u

= \int \frac{1}{3 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{3} \cdot \int \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{3} tan (u)

Substitute back

u = arcsin (3 x)

= \frac{1}{3} tan (arcsin (3 x))

Simplify

\frac{1}{3} tan (arcsin (3 x)) : \frac{x}{( 1 - 9 x ^{2} ) ^{\frac{1}{2}}}

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

Add a constant to the solution

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

\int \frac{1}{3} x^{5} d x

\int + 9 ln (x^{2} - 1) d x

\int \frac{x ^{3} + 4 x + 3}{x ^{2} + 2} d x

\int x (2 x^{2} + 3)^{6} d x

\int \frac{( ln ( x ) - 2 ) ^{3} ( ln ( x ) + 1 )}{x} d x

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