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

The integral of (x^2)/(x^6+1) is 1/3 arctan(x^3)+C

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

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

\frac{1}{3} arctan (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

= \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 = arctan (u)

= \frac{1}{3} arctan (u)

Substitute back

u = x^{3}

= \frac{1}{3} arctan (x^{3})

Add a constant to the solution

= \frac{1}{3} arctan (x^{3}) + C

f (x) = x^{3} - x

\frac{d}{d x} ((x^{2} - 8) (5 x^{3}))

\int \frac{cos ( ln ( 2 x ) )}{x} d x

\int \frac{x ^{2} + 16 x - 4}{x ^{3} - 4 x} d x

y = \frac{1}{3 + 2 x}, (2, \frac{1}{7})

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