What is the integral of (x^2)/(sqrt(x^6+1)) ?

The integral of (x^2)/(sqrt(x^6+1)) is 1/3 ln|sqrt(x^6+1)+x^3|+C

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integral of (x^2)/(sqrt(x^6+1))

\int \frac{x ^{2}}{x ^{6} + 1} d x

\frac{1}{3} ln x^{6} + 1 + x^{3} + C

Solution steps

\int \frac{x ^{2}}{x ^{6} + 1} d x

Apply u-substitution

= \int \frac{1}{3 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}{3} \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}{3} ln u^{2} + 1 + u

Substitute back

u = x^{3}

= \frac{1}{3} ln (x^{3})^{2} + 1 + x^{3}

(x^{3})^{2} = x^{6}

= \frac{1}{3} ln x^{6} + 1 + x^{3}

Add a constant to the solution

= \frac{1}{3} ln x^{6} + 1 + x^{3} + C

What is the integral of (x^2)/(sqrt(x^6+1)) ?
The integral of (x^2)/(sqrt(x^6+1)) is 1/3 ln|sqrt(x^6+1)+x^3|+C