What is the integral of (2v^2)/((1-v^2)^{5/2)} ?

The integral of (2v^2)/((1-v^2)^{5/2)} is (2v^3)/(3(1-v^2)^{3/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2v^2)/((1-v^2)^{5/2)}

\int \frac{2 v ^{2}}{( 1 - v ^{2} ) ^{\frac{5}{2}}} d v

\frac{2 v ^{3}}{3 ( 1 - v ^{2} ) ^{\frac{3}{2}}} + C

Solution steps

\int \frac{2 v ^{2}}{( 1 - v ^{2} ) ^{\frac{5}{2}}} d v

Take the constant out:

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

= 2 \cdot \int \frac{v ^{2}}{( 1 - v ^{2} ) ^{\frac{5}{2}}} d v

Apply Trigonometric Substitution

: \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

= 2 \cdot \int \frac{sin ^{2} ( u )}{cos ^{4} ( u )} d u

Rewrite using trig identities

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

Apply u-substitution

: \int w^{2} d w

= 2 \cdot \int w^{2} d w

Apply the Power Rule

: \frac{w ^{3}}{3}

= 2 \cdot \frac{w ^{3}}{3}

Substitute back

= 2 \cdot \frac{tan ^{3} ( arcsin ( v ) )}{3}

Simplify

2 \cdot \frac{tan ^{3} ( arcsin ( v ) )}{3} : \frac{2 v ^{3}}{3 ( 1 - v ^{2} ) ^{\frac{3}{2}}}

= \frac{2 v ^{3}}{3 ( 1 - v ^{2} ) ^{\frac{3}{2}}}

Add a constant to the solution

= \frac{2 v ^{3}}{3 ( 1 - v ^{2} ) ^{\frac{3}{2}}} + C

\int \frac{- 2 x ^{2}}{4 x - x ^{2}} d x

\int 2 π x sin (x) d x

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

\int 2^{4 x} d x

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

What is the integral of (2v^2)/((1-v^2)^{5/2)} ?
The integral of (2v^2)/((1-v^2)^{5/2)} is (2v^3)/(3(1-v^2)^{3/2)}+C