What is the integral of (2x+1)/(x^2+2x+26) ?

The integral of (2x+1)/(x^2+2x+26) is ln|x^2+2x+26|-1/5 arctan(1/5 (x+1))+C

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integral of (2x+1)/(x^2+2x+26)

\int \frac{2 x + 1}{x ^{2} + 2 x + 26} d x

ln x^{2} + 2 x + 26 - \frac{1}{5} arctan (\frac{1}{5} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 26} d x

Complete the square

x^{2} + 2 x + 26 : (x + 1)^{2} + 25

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 25} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 25} d u

= \int \frac{2 u - 1}{u ^{2} + 25} d u

Expand

\frac{2 u - 1}{u ^{2} + 25} : \frac{2 u}{u ^{2} + 25} - \frac{1}{u ^{2} + 25}

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} + 25} d u - \int \frac{1}{u ^{2} + 25} d u

\int \frac{2 u}{u ^{2} + 25} d u = ln u^{2} + 25

\int \frac{1}{u ^{2} + 25} d u = \frac{1}{5} arctan (\frac{u}{5})

= ln u^{2} + 25 - \frac{1}{5} arctan (\frac{u}{5})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 25 - \frac{1}{5} arctan (\frac{x + 1}{5})

Simplify

ln (x + 1)^{2} + 25 - \frac{1}{5} arctan (\frac{x + 1}{5}) : ln x^{2} + 2 x + 26 - \frac{1}{5} arctan (\frac{1}{5} (x + 1))

= ln x^{2} + 2 x + 26 - \frac{1}{5} arctan (\frac{1}{5} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 26 - \frac{1}{5} arctan (\frac{1}{5} (x + 1)) + C

\int sec^{n} (x) tan (x) d x

\frac{d}{d x} (1 - (1 - \frac{x}{24})^{3})

y^{^{'}} + 3 y = 6

f (x) = (x - 5)^{6} (x + 3)^{2}

\frac{d}{d x} (sin^{\frac{1}{2}} (2 x))

What is the integral of (2x+1)/(x^2+2x+26) ?
The integral of (2x+1)/(x^2+2x+26) is ln|x^2+2x+26|-1/5 arctan(1/5 (x+1))+C