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

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

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

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

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

Solution steps

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

Apply u-substitution

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

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

Use the common integral:

\int \frac{1}{u ^{2} + 1} 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 x^{- \frac{3}{5}} d x

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\int \frac{1}{x ln ( 5 x )} d x

\int (sec (x))^{3} tan (x) d x

2 y^{3}

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