What is the integral of x/(4x^2+2x+5) ?

The integral of x/(4x^2+2x+5) is 1/8 ln|16x^2+8x+20|-1/(4sqrt(19))arctan((4x+1)/(sqrt(19)))+C

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integral of x/(4x^2+2x+5)

\int \frac{x}{4 x ^{2} + 2 x + 5} d x

\frac{1}{8} ln 16 x^{2} + 8 x + 20 - \frac{1}{4 19} arctan (\frac{4 x + 1}{19}) + C

Solution steps

\int \frac{x}{4 x ^{2} + 2 x + 5} d x

Complete the square

4 x^{2} + 2 x + 5 : 4 (x + \frac{1}{4})^{2} + \frac{19}{4}

= \int \frac{x}{4 ( x + \frac{1}{4} ) ^{2} + \frac{19}{4}} d x

Apply u-substitution

: \int \frac{4 u - 1}{16 u ^{2} + 19} d u

= \int \frac{4 u - 1}{16 u ^{2} + 19} d u

Expand

\frac{4 u - 1}{16 u ^{2} + 19} : \frac{4 u}{16 u ^{2} + 19} - \frac{1}{16 u ^{2} + 19}

Apply the Sum Rule:

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

= \int \frac{4 u}{16 u ^{2} + 19} d u - \int \frac{1}{16 u ^{2} + 19} d u

\int \frac{4 u}{16 u ^{2} + 19} d u = \frac{1}{8} ln 16 u^{2} + 19

\int \frac{1}{16 u ^{2} + 19} d u = \frac{1}{4 19} arctan (\frac{4}{19} u)

= \frac{1}{8} ln 16 u^{2} + 19 - \frac{1}{4 19} arctan (\frac{4}{19} u)

Substitute back

u = x + \frac{1}{4}

= \frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 19 - \frac{1}{4 19} arctan (\frac{4}{19} (x + \frac{1}{4}))

Simplify

\frac{1}{8} ln 16 (x + \frac{1}{4})^{2} + 19 - \frac{1}{4 19} arctan (\frac{4}{19} (x + \frac{1}{4})) : \frac{1}{8} ln 16 x^{2} + 8 x + 20 - \frac{1}{4 19} arctan (\frac{4 x + 1}{19})

= \frac{1}{8} ln 16 x^{2} + 8 x + 20 - \frac{1}{4 19} arctan (\frac{4 x + 1}{19})

Add a constant to the solution

= \frac{1}{8} ln 16 x^{2} + 8 x + 20 - \frac{1}{4 19} arctan (\frac{4 x + 1}{19}) + C

\int - 2 x^{9} d x

2 x d x

y = \frac{t}{( t - 8 ) ^{2}}

\int_{0}^{\frac{π}{2}} x sin (\frac{x}{3}) d x

\frac{\partial}{\partial s} (- 2 s^{2} + t^{2})

What is the integral of x/(4x^2+2x+5) ?
The integral of x/(4x^2+2x+5) is 1/8 ln|16x^2+8x+20|-1/(4sqrt(19))arctan((4x+1)/(sqrt(19)))+C