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

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

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integral of (3x^2)/(1+x^6)

\int \frac{3 x ^{2}}{1 + x ^{6}} d x

arctan (x^{3}) + C

Solution steps

\int \frac{3 x ^{2}}{1 + x ^{6}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int \frac{x ^{2}}{1 + x ^{6}} d x

Apply u-substitution

: \int \frac{1}{3 ( 1 + u ^{2} )} d u

= 3 \cdot \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

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

Substitute back

u = x^{3}

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

Simplify

3 \cdot \frac{1}{3} arctan (x^{3}) : arctan (x^{3})

= arctan (x^{3})

Add a constant to the solution

= arctan (x^{3}) + C

\int x^{2} cos (x^{3} - 1) d x

sin (x^{2})

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\int (\frac{5 x - 2}{( x + 2 ) ^{2}}) d x

\int \frac{x ^{2}}{( 1 - x ^{2} ) ^{3}} d x

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