What is the integral of (34)/((1-x^2)^{3/2)} ?

The integral of (34)/((1-x^2)^{3/2)} is (34x)/((1-x^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 (34)/((1-x^2)^{3/2)}

\int \frac{34}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

\frac{34 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

Solution steps

\int \frac{34}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Take the constant out:

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

= 34 \cdot \int \frac{1}{( 1 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

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

= 34 \cdot \int sec^{2} (u) d u

Use the common integral:

\int sec^{2} (u) d u = tan (u)

= 34 tan (u)

Substitute back

u = arcsin (x)

= 34 tan (arcsin (x))

Simplify

34 tan (arcsin (x)) : \frac{34 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

= \frac{34 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{34 x}{( 1 - x ^{2} ) ^{\frac{1}{2}}} + C

\int \frac{5 x ^{2} - 3 x + 2}{( x - 1 ) ( x ^{2} + 1 )} d x

\int (3 x^{\frac{1}{2}} + x^{4}) d x

\int 7 sec^{3} (x) d x

\int 9 x tan^{2} (3 x) d x

\int \frac{9 sin ( x )}{3 cos ( x )} d x

What is the integral of (34)/((1-x^2)^{3/2)} ?
The integral of (34)/((1-x^2)^{3/2)} is (34x)/((1-x^2)^{1/2)}+C