What is the integral of x/(4x^2+32x+70) ?

The integral of x/(4x^2+32x+70) is 1/8 ln|2x^2+16x+35|-(sqrt(2))/(sqrt(3))arctan(sqrt(2/3)(x+4))+C

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integral of x/(4x^2+32x+70)

\int \frac{x}{4 x ^{2} + 32 x + 70} d x

\frac{1}{8} ln 2 x^{2} + 16 x + 35 - \frac{2}{3} arctan (\frac{2}{3} (x + 4)) + C

Solution steps

\int \frac{x}{4 x ^{2} + 32 x + 70} d x

\frac{x}{4 x ^{2} + 32 x + 70} = \frac{1}{2} \frac{x}{2 x ^{2} + 16 x + 35}

= \int \frac{1}{2} \cdot \frac{x}{2 x ^{2} + 16 x + 35} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{x}{2 x ^{2} + 16 x + 35} d x

Complete the square

2 x^{2} + 16 x + 35 : 2 (x + 4)^{2} + 3

= \frac{1}{2} \cdot \int \frac{x}{2 ( x + 4 ) ^{2} + 3} d x

Apply u-substitution

: \int \frac{u - 4}{2 u ^{2} + 3} d u

= \frac{1}{2} \cdot \int \frac{u - 4}{2 u ^{2} + 3} d u

Expand

\frac{u - 4}{2 u ^{2} + 3} : \frac{u}{2 u ^{2} + 3} - \frac{4}{2 u ^{2} + 3}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} (\int \frac{u}{2 u ^{2} + 3} d u - \int \frac{4}{2 u ^{2} + 3} d u)

\int \frac{u}{2 u ^{2} + 3} d u = \frac{1}{4} ln 2 u^{2} + 3

\int \frac{4}{2 u ^{2} + 3} d u = \frac{2 2}{3} arctan (\frac{2}{3} u)

= \frac{1}{2} (\frac{1}{4} ln 2 u^{2} + 3 - \frac{2 2}{3} arctan (\frac{2}{3} u))

Substitute back

u = x + 4

= \frac{1}{2} (\frac{1}{4} ln 2 (x + 4)^{2} + 3 - \frac{2 2}{3} arctan (\frac{2}{3} (x + 4)))

Simplify

\frac{1}{2} (\frac{1}{4} ln 2 (x + 4)^{2} + 3 - \frac{2 2}{3} arctan (\frac{2}{3} (x + 4))) : \frac{1}{8} ln 2 x^{2} + 16 x + 35 - \frac{2}{3} arctan (\frac{2}{3} (x + 4))

= \frac{1}{8} ln 2 x^{2} + 16 x + 35 - \frac{2}{3} arctan (\frac{2}{3} (x + 4))

Add a constant to the solution

= \frac{1}{8} ln 2 x^{2} + 16 x + 35 - \frac{2}{3} arctan (\frac{2}{3} (x + 4)) + C

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What is the integral of x/(4x^2+32x+70) ?
The integral of x/(4x^2+32x+70) is 1/8 ln|2x^2+16x+35|-(sqrt(2))/(sqrt(3))arctan(sqrt(2/3)(x+4))+C