What is the integral of (2x+1)/(x^2+2x+65) ?

The integral of (2x+1)/(x^2+2x+65) is ln|x^2+2x+65|-1/8 arctan(1/8 (x+1))+C

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integral of (2x+1)/(x^2+2x+65)

\int \frac{2 x + 1}{x ^{2} + 2 x + 65} d x

ln x^{2} + 2 x + 65 - \frac{1}{8} arctan (\frac{1}{8} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 65} d x

Complete the square

x^{2} + 2 x + 65 : (x + 1)^{2} + 64

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 64} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 64} d u

= \int \frac{2 u - 1}{u ^{2} + 64} d u

Expand

\frac{2 u - 1}{u ^{2} + 64} : \frac{2 u}{u ^{2} + 64} - \frac{1}{u ^{2} + 64}

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} + 64} d u - \int \frac{1}{u ^{2} + 64} d u

\int \frac{2 u}{u ^{2} + 64} d u = ln u^{2} + 64

\int \frac{1}{u ^{2} + 64} d u = \frac{1}{8} arctan (\frac{u}{8})

= ln u^{2} + 64 - \frac{1}{8} arctan (\frac{u}{8})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 64 - \frac{1}{8} arctan (\frac{x + 1}{8})

Simplify

ln (x + 1)^{2} + 64 - \frac{1}{8} arctan (\frac{x + 1}{8}) : ln x^{2} + 2 x + 65 - \frac{1}{8} arctan (\frac{1}{8} (x + 1))

= ln x^{2} + 2 x + 65 - \frac{1}{8} arctan (\frac{1}{8} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 65 - \frac{1}{8} arctan (\frac{1}{8} (x + 1)) + C

\frac{d}{d x} (\frac{3 x ^{2} + 1}{2 x})

\int sin^{3} (b) cos^{3} (b)

x \to - 2 lim (x^{4} + 5^{x} - 2)

\frac{\partial}{\partial y} (x ln (x + 2 y + 3 z))

\frac{\partial}{\partial x} (\frac{x ^{3}}{3})

What is the integral of (2x+1)/(x^2+2x+65) ?
The integral of (2x+1)/(x^2+2x+65) is ln|x^2+2x+65|-1/8 arctan(1/8 (x+1))+C