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

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

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

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

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

Solution steps

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

Apply u-substitution

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

= \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

= \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

= \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

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