What is the integral of (5xsqrt(x))/(1+x^5) ?

The integral of (5xsqrt(x))/(1+x^5) is 2arctan(x^{5/2})+C

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integral of (5xsqrt(x))/(1+x^5)

\int \frac{5 x x}{1 + x ^{5}} d x

2 arctan (x^{\frac{5}{2}}) + C

Solution steps

\int \frac{5 x x}{1 + x ^{5}} d x

Take the constant out:

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

= 5 \cdot \int \frac{x x}{1 + x ^{5}} d x

Apply u-substitution

: \int \frac{2 u ^{4}}{1 + u ^{10}} d u

= 5 \cdot \int \frac{2 u ^{4}}{1 + u ^{10}} d u

Take the constant out:

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

= 5 \cdot 2 \cdot \int \frac{u ^{4}}{1 + u ^{10}} d u

Apply u-substitution

: \int \frac{1}{5 ( 1 + v ^{2} )} d v

= 5 \cdot 2 \cdot \int \frac{1}{5 ( 1 + v ^{2} )} d v

Take the constant out:

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

= 5 \cdot 2 \cdot \frac{1}{5} \cdot \int \frac{1}{1 + v ^{2}} d v

Use the common integral:

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

= 5 \cdot 2 \cdot \frac{1}{5} arctan (v)

Substitute back

= 5 \cdot 2 \cdot \frac{1}{5} arctan ((x)^{5})

Simplify

5 \cdot 2 \cdot \frac{1}{5} arctan ((x)^{5}) : 2 arctan (x^{\frac{5}{2}})

= 2 arctan (x^{\frac{5}{2}})

Add a constant to the solution

= 2 arctan (x^{\frac{5}{2}}) + C

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What is the integral of (5xsqrt(x))/(1+x^5) ?
The integral of (5xsqrt(x))/(1+x^5) is 2arctan(x^{5/2})+C