What is the integral of (x^2)/(sqrt(1-4x^6)) ?

The integral of (x^2)/(sqrt(1-4x^6)) is 1/6 arcsin(2x^3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x^2)/(sqrt(1-4x^6))

\int \frac{x ^{2}}{1 - 4 x ^{6}} d x

\frac{1}{6} arcsin (2 x^{3}) + C

Solution steps

\int \frac{x ^{2}}{1 - 4 x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 1 - 4 u ^{2}} d u

= \int \frac{1}{3 1 - 4 u ^{2}} 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}{1 - 4 u ^{2}} d u

Apply Trigonometric Substitution

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

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

Take the constant out:

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

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

Integral of a constant:

\int a d x = a x

= \frac{1}{3} \cdot \frac{1}{2} \cdot 1 \cdot v

Substitute back

= \frac{1}{3} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{3})

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot 1 \cdot arcsin (2 x^{3}) : \frac{1}{6} arcsin (2 x^{3})

= \frac{1}{6} arcsin (2 x^{3})

Add a constant to the solution

= \frac{1}{6} arcsin (2 x^{3}) + C

\int x 1 - 9 x^{2} d x

\int \frac{x - 2}{x ^{2} - 2 x + 5} d x

\int \frac{sin ( 2 x )}{21 + cos ^{2} ( x )} d x

\int x^{8} (x^{9} - 4)^{4} d x

\int \frac{9 x ^{3} - 9 x + 2}{x ^{3} - x ^{2}} d x

What is the integral of (x^2)/(sqrt(1-4x^6)) ?
The integral of (x^2)/(sqrt(1-4x^6)) is 1/6 arcsin(2x^3)+C