What is the integral of x/(25+16x^4) ?

The integral of x/(25+16x^4) is 1/40 arctan(4/5 x^2)+C

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integral of x/(25+16x^4)

\int \frac{x}{25 + 16 x ^{4}} d x

\frac{1}{40} arctan (\frac{4}{5} x^{2}) + C

Solution steps

\int \frac{x}{25 + 16 x ^{4}} d x

Apply u-substitution

: \int \frac{1}{2 ( 25 + 16 u ^{2} )} d u

= \int \frac{1}{2 ( 25 + 16 u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{25 + 16 u ^{2}} d u

Apply Integral Substitution

= \frac{1}{2} \cdot \int \frac{1}{20 ( v ^{2} + 1 )} d v

Take the constant out:

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

= \frac{1}{2} \cdot \frac{1}{20} \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{2} \cdot \frac{1}{20} arctan (v)

Substitute back

= \frac{1}{2} \cdot \frac{1}{20} arctan (\frac{4}{5} x^{2})

Simplify

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

= \frac{1}{40} arctan (\frac{4}{5} x^{2})

Add a constant to the solution

= \frac{1}{40} arctan (\frac{4}{5} x^{2}) + C

\int \frac{x ^{2}}{6 - x} d x

x \to 2 lim (f (x))

y = x^{3} - 9 x^{2} + 18 x, y = - x^{3} + 9 x^{2} - 18 x

\int x sin (6 - x) d x

\int \frac{16}{x ^{2} + 1} d x

What is the integral of x/(25+16x^4) ?
The integral of x/(25+16x^4) is 1/40 arctan(4/5 x^2)+C