What is the integral of x/(x^4+8x^2+17) ?

The integral of x/(x^4+8x^2+17) is 1/2 arctan(x^2+4)+C

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integral of x/(x^4+8x^2+17)

\int \frac{x}{x ^{4} + 8 x ^{2} + 17} d x

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

Solution steps

\int \frac{x}{x ^{4} + 8 x ^{2} + 17} d x

Apply u-substitution

: \int \frac{1}{4 u ( 8 u + u + 17 )} d u

= \int \frac{1}{4 u ( 8 u + u + 17 )} 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 ( 8 u + u + 17 )} d u

Apply u-substitution

: \int \frac{2}{8 v + v ^{2} + 17} d v

= \frac{1}{4} \cdot \int \frac{2}{8 v + v ^{2} + 17} 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}{8 v + v ^{2} + 17} d v

Complete the square

8 v + v^{2} + 17 : (v + 4)^{2} + 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (y cos (x))

\int_{5}^{e^{3}} \frac{1}{x} d x

y^{^{'}} = x e^{y}

\frac{\partial}{\partial y} (sin^{2} (y))

\frac{\partial}{\partial x} (x^{e^{x}})

What is the integral of x/(x^4+8x^2+17) ?
The integral of x/(x^4+8x^2+17) is 1/2 arctan(x^2+4)+C