What is the integral of x/(x^4+10x^2+26) ?

The integral of x/(x^4+10x^2+26) is 1/2 arctan(x^2+5)+C

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integral of x/(x^4+10x^2+26)

\int \frac{x}{x ^{4} + 10 x ^{2} + 26} d x

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

Solution steps

\int \frac{x}{x ^{4} + 10 x ^{2} + 26} d x

Apply u-substitution

: \int \frac{1}{4 u ( 10 u + u + 26 )} d u

= \int \frac{1}{4 u ( 10 u + u + 26 )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int \frac{1}{u ( 10 u + u + 26 )} d u

Apply u-substitution

: \int \frac{2}{10 v + v ^{2} + 26} d v

= \frac{1}{4} \cdot \int \frac{2}{10 v + v ^{2} + 26} d v

Take the constant out:

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

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{10 v + v ^{2} + 26} d v

Complete the square

10 v + v^{2} + 26 : (v + 5)^{2} + 1

= \frac{1}{4} \cdot 2 \cdot \int \frac{1}{( v + 5 ) ^{2} + 1} d v

Apply u-substitution

: \int \frac{1}{w ^{2} + 1} d w

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

Use the common integral:

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

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

x \to 0 lim ((\frac{2}{3})^{x})

\int sin (3 x) \cdot sin (5 x) d x

x \to 0 lim (- \frac{2}{x})

\frac{d}{d x} (\frac{1}{5 x - 1})

arcsin (x^{4})

What is the integral of x/(x^4+10x^2+26) ?
The integral of x/(x^4+10x^2+26) is 1/2 arctan(x^2+5)+C