What is the integral of (2x+1)/(x^2+2x+37) ?

The integral of (2x+1)/(x^2+2x+37) is ln|x^2+2x+37|-1/6 arctan(1/6 (x+1))+C

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integral of (2x+1)/(x^2+2x+37)

\int \frac{2 x + 1}{x ^{2} + 2 x + 37} d x

ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 37} d x

Complete the square

x^{2} + 2 x + 37 : (x + 1)^{2} + 36

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 36} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 36} d u

= \int \frac{2 u - 1}{u ^{2} + 36} d u

Expand

\frac{2 u - 1}{u ^{2} + 36} : \frac{2 u}{u ^{2} + 36} - \frac{1}{u ^{2} + 36}

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{2 u}{u ^{2} + 36} d u - \int \frac{1}{u ^{2} + 36} d u

\int \frac{2 u}{u ^{2} + 36} d u = ln u^{2} + 36

\int \frac{1}{u ^{2} + 36} d u = \frac{1}{6} arctan (\frac{u}{6})

= ln u^{2} + 36 - \frac{1}{6} arctan (\frac{u}{6})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 36 - \frac{1}{6} arctan (\frac{x + 1}{6})

Simplify

ln (x + 1)^{2} + 36 - \frac{1}{6} arctan (\frac{x + 1}{6}) : ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1))

= ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 37 - \frac{1}{6} arctan (\frac{1}{6} (x + 1)) + C

\frac{\partial}{\partial z} (x^{2} yz)

\frac{d}{d x} (sin (x) \cdot sin (x))

\int_{- 9}^{9} 81 - x^{2} d x

t y^{^{'}} - 2 y = t^{5} sin (2 t) - t^{3} + 4 t^{4}

f (x) = \frac{12 x}{x ^{2} - 4}, at x = 4

What is the integral of (2x+1)/(x^2+2x+37) ?
The integral of (2x+1)/(x^2+2x+37) is ln|x^2+2x+37|-1/6 arctan(1/6 (x+1))+C