What is the integral of (2x+1)/(x^2+2x+50) ?

The integral of (2x+1)/(x^2+2x+50) is ln|x^2+2x+50|-1/7 arctan(1/7 (x+1))+C

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integral of (2x+1)/(x^2+2x+50)

\int \frac{2 x + 1}{x ^{2} + 2 x + 50} d x

ln x^{2} + 2 x + 50 - \frac{1}{7} arctan (\frac{1}{7} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 50} d x

Complete the square

x^{2} + 2 x + 50 : (x + 1)^{2} + 49

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 49} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 49} d u

= \int \frac{2 u - 1}{u ^{2} + 49} d u

Expand

\frac{2 u - 1}{u ^{2} + 49} : \frac{2 u}{u ^{2} + 49} - \frac{1}{u ^{2} + 49}

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} + 49} d u - \int \frac{1}{u ^{2} + 49} d u

\int \frac{2 u}{u ^{2} + 49} d u = ln u^{2} + 49

\int \frac{1}{u ^{2} + 49} d u = \frac{1}{7} arctan (\frac{u}{7})

= ln u^{2} + 49 - \frac{1}{7} arctan (\frac{u}{7})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 49 - \frac{1}{7} arctan (\frac{x + 1}{7})

Simplify

ln (x + 1)^{2} + 49 - \frac{1}{7} arctan (\frac{x + 1}{7}) : ln x^{2} + 2 x + 50 - \frac{1}{7} arctan (\frac{1}{7} (x + 1))

= ln x^{2} + 2 x + 50 - \frac{1}{7} arctan (\frac{1}{7} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 50 - \frac{1}{7} arctan (\frac{1}{7} (x + 1)) + C

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

\frac{d y}{d x}, ln (y) = x

\int \frac{y}{1} d y

x^{(\frac{1}{3})}

\int \frac{1}{1 - v ^{2}} d v

What is the integral of (2x+1)/(x^2+2x+50) ?
The integral of (2x+1)/(x^2+2x+50) is ln|x^2+2x+50|-1/7 arctan(1/7 (x+1))+C