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

The integral of (5x)/(sqrt(1+x^4)) is 5/2 ln|sqrt(x^4+1)+x^2|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (5x)/(sqrt(1+x^4))

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

\frac{5}{2} ln x^{4} + 1 + x^{2} + C

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

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

= 5 \cdot \frac{1}{2} ln u^{2} + 1 + u

Substitute back

u = x^{2}

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

Simplify

5 \cdot \frac{1}{2} ln (x^{2})^{2} + 1 + x^{2} : \frac{5}{2} ln x^{4} + 1 + x^{2}

= \frac{5}{2} ln x^{4} + 1 + x^{2}

Add a constant to the solution

= \frac{5}{2} ln x^{4} + 1 + x^{2} + C

\int y \cdot sin (y) d y

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

\int e^{x} cot^{2} (e^{x} + 1) d x

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

\int \frac{( ln ^{4} ( x ) )}{x} d x

What is the integral of (5x)/(sqrt(1+x^4)) ?
The integral of (5x)/(sqrt(1+x^4)) is 5/2 ln|sqrt(x^4+1)+x^2|+C