What is the integral of (10x)/(4+25x^4) ?

The integral of (10x)/(4+25x^4) is 1/2 arctan(5/2 x^2)+C

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integral of (10x)/(4+25x^4)

\int \frac{10 x}{4 + 25 x ^{4}} d x

\frac{1}{2} arctan (\frac{5}{2} x^{2}) + C

Solution steps

\int \frac{10 x}{4 + 25 x ^{4}} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{4 + 25 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 4 + 25 u ^{2} )} d u

= 10 \cdot \int \frac{1}{2 ( 4 + 25 u ^{2} )} d u

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{4 + 25 u ^{2}} d u

Apply Integral Substitution

= 10 \cdot \frac{1}{2} \cdot \int \frac{1}{10 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 10 \cdot \frac{1}{2} \cdot \frac{1}{10} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 10 \cdot \frac{1}{2} \cdot \frac{1}{10} arctan (v)

Substitute back

= 10 \cdot \frac{1}{2} \cdot \frac{1}{10} arctan (\frac{5}{2} x^{2})

Simplify

10 \cdot \frac{1}{2} \cdot \frac{1}{10} arctan (\frac{5}{2} x^{2}) : \frac{1}{2} arctan (\frac{5}{2} x^{2})

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

Add a constant to the solution

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

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What is the integral of (10x)/(4+25x^4) ?
The integral of (10x)/(4+25x^4) is 1/2 arctan(5/2 x^2)+C