What is the integral of (7x^2)/(1+x^6) ?

The integral of (7x^2)/(1+x^6) is 7/3 arctan(x^3)+C

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integral of (7x^2)/(1+x^6)

\int \frac{7 x ^{2}}{1 + x ^{6}} d x

\frac{7}{3} arctan (x^{3}) + C

Solution steps

\int \frac{7 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

= 7 \cdot \int \frac{x ^{2}}{1 + x ^{6}} d x

Apply u-substitution

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

= 7 \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

= 7 \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)

= 7 \cdot \frac{1}{3} arctan (u)

Substitute back

u = x^{3}

= 7 \cdot \frac{1}{3} arctan (x^{3})

Simplify

7 \cdot \frac{1}{3} arctan (x^{3}) : \frac{7}{3} arctan (x^{3})

= \frac{7}{3} arctan (x^{3})

Add a constant to the solution

= \frac{7}{3} arctan (x^{3}) + C

(x^{2} y^{2} - x^{4}) y^{^{'}} = x^{2} y^{2} - y^{4}

x \to - 1 lim (ln (1 + x))

\frac{d}{d x} (ln (4 x - 5))

\frac{1}{1 - x - x ^{2}}

t - 2

What is the integral of (7x^2)/(1+x^6) ?
The integral of (7x^2)/(1+x^6) is 7/3 arctan(x^3)+C