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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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

= \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

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

Substitute back

u = x^{2}

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

(x^{2})^{2} = x^{4}

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

Add a constant to the solution

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

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