What is the integral of (4x)/(100+16x^4) ?

The integral of (4x)/(100+16x^4) is 1/20 arctan(2/5 x^2)+C

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integral of (4x)/(100+16x^4)

\int \frac{4 x}{100 + 16 x ^{4}} d x

\frac{1}{20} arctan (\frac{2}{5} x^{2}) + C

Solution steps

\int \frac{4 x}{100 + 16 x ^{4}} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{100 + 16 x ^{4}} d x

Apply u-substitution

= 4 \cdot \int \frac{1}{2 ( 100 + 16 u ^{2} )} d u

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{100 + 16 u ^{2}} d u

Apply Integral Substitution

= 4 \cdot \frac{1}{2} \cdot \int \frac{1}{40 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \frac{1}{40} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= 4 \cdot \frac{1}{2} \cdot \frac{1}{40} arctan (v)

Substitute back

= 4 \cdot \frac{1}{2} \cdot \frac{1}{40} arctan (\frac{4}{10} x^{2})

Simplify

4 \cdot \frac{1}{2} \cdot \frac{1}{40} arctan (\frac{4}{10} x^{2}) : \frac{1}{20} arctan (\frac{2}{5} x^{2})

= \frac{1}{20} arctan (\frac{2}{5} x^{2})

Add a constant to the solution

= \frac{1}{20} arctan (\frac{2}{5} x^{2}) + C

What is the integral of (4x)/(100+16x^4) ?
The integral of (4x)/(100+16x^4) is 1/20 arctan(2/5 x^2)+C