What is the integral of (5x^2)/(1+4x^6) ?

The integral of (5x^2)/(1+4x^6) is 5/6 arctan(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 (5x^2)/(1+4x^6)

\int \frac{5 x ^{2}}{1 + 4 x ^{6}} d x

\frac{5}{6} arctan (2 x^{3}) + C

Solution steps

\int \frac{5 x ^{2}}{1 + 4 x ^{6}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x ^{2}}{1 + 4 x ^{6}} d x

Apply u-substitution

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

= 5 \cdot \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

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{1 + 4 u ^{2}} d u

Apply Integral Substitution

= 5 \cdot \frac{1}{3} \cdot \int \frac{1}{2 ( 1 + v ^{2} )} d v

Take the constant out:

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

= 5 \cdot \frac{1}{3} \cdot \frac{1}{2} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

\int \frac{1}{1 + v ^{2}} d v = arctan (v)

= 5 \cdot \frac{1}{3} \cdot \frac{1}{2} arctan (v)

Substitute back

= 5 \cdot \frac{1}{3} \cdot \frac{1}{2} arctan (2 x^{3})

Simplify

5 \cdot \frac{1}{3} \cdot \frac{1}{2} arctan (2 x^{3}) : \frac{5}{6} arctan (2 x^{3})

= \frac{5}{6} arctan (2 x^{3})

Add a constant to the solution

= \frac{5}{6} arctan (2 x^{3}) + C

\int x \cdot x - 2 d x

\int - \frac{1}{8} e^{- 2 x} d x

\int (\frac{sin ( 2 x )}{1 + cos ^{2} ( x )}) d x

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

What is the integral of (5x^2)/(1+4x^6) ?
The integral of (5x^2)/(1+4x^6) is 5/6 arctan(2x^3)+C