What is the integral of x/(1+25x^4) ?

The integral of x/(1+25x^4) is 1/10 arctan(5x^2)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of x/(1+25x^4)

\int \frac{x}{1 + 25 x ^{4}} d x

\frac{1}{10} arctan (5 x^{2}) + C

Solution steps

\int \frac{x}{1 + 25 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 1 + 25 u ^{2} )} d u

= \int \frac{1}{2 ( 1 + 25 u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{1 + 25 u ^{2}} d u

Apply Integral Substitution

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

Take the constant out:

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

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

Use the common integral:

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

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

Substitute back

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

Simplify

\frac{1}{2} \cdot \frac{1}{5} arctan (5 x^{2}) : \frac{1}{10} arctan (5 x^{2})

= \frac{1}{10} arctan (5 x^{2})

Add a constant to the solution

= \frac{1}{10} arctan (5 x^{2}) + C

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

x \to \frac{π}{3} lim (sec (x))

x \to π lim (0.4)

y = tan (5 x^{3})

\frac{\partial}{\partial x} (x \cdot cos (x))

What is the integral of x/(1+25x^4) ?
The integral of x/(1+25x^4) is 1/10 arctan(5x^2)+C