What is the integral of (2x+1)/(x^2+2x+122) ?

The integral of (2x+1)/(x^2+2x+122) is ln|x^2+2x+122|-1/11 arctan(1/11 (x+1))+C

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integral of (2x+1)/(x^2+2x+122)

\int \frac{2 x + 1}{x ^{2} + 2 x + 122} d x

ln x^{2} + 2 x + 122 - \frac{1}{11} arctan (\frac{1}{11} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 122} d x

Complete the square

x^{2} + 2 x + 122 : (x + 1)^{2} + 121

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 121} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 121} d u

= \int \frac{2 u - 1}{u ^{2} + 121} d u

Expand

\frac{2 u - 1}{u ^{2} + 121} : \frac{2 u}{u ^{2} + 121} - \frac{1}{u ^{2} + 121}

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} + 121} d u - \int \frac{1}{u ^{2} + 121} d u

\int \frac{2 u}{u ^{2} + 121} d u = ln u^{2} + 121

\int \frac{1}{u ^{2} + 121} d u = \frac{1}{11} arctan (\frac{u}{11})

= ln u^{2} + 121 - \frac{1}{11} arctan (\frac{u}{11})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 121 - \frac{1}{11} arctan (\frac{x + 1}{11})

Simplify

ln (x + 1)^{2} + 121 - \frac{1}{11} arctan (\frac{x + 1}{11}) : ln x^{2} + 2 x + 122 - \frac{1}{11} arctan (\frac{1}{11} (x + 1))

= ln x^{2} + 2 x + 122 - \frac{1}{11} arctan (\frac{1}{11} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 122 - \frac{1}{11} arctan (\frac{1}{11} (x + 1)) + C

\frac{\partial}{\partial y} (x^{2} y + y)

\int_{0}^{u} v + x d v

(1 - t^{2}) y^{^{'}} - t^{2} y = (1 + t) \cdot (1 - t^{2})

3 x^{2} + y^{2} = 1, (\frac{1}{2}, \frac{- 1}{2})

y^{^{''}} - y^{^{'}} + 12 y = 0

What is the integral of (2x+1)/(x^2+2x+122) ?
The integral of (2x+1)/(x^2+2x+122) is ln|x^2+2x+122|-1/11 arctan(1/11 (x+1))+C