What is the integral of (2x+1)/(x^2+2x+10) ?

The integral of (2x+1)/(x^2+2x+10) is ln|x^2+2x+10|-1/3 arctan(1/3 (x+1))+C

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integral of (2x+1)/(x^2+2x+10)

\int \frac{2 x + 1}{x ^{2} + 2 x + 10} d x

ln x^{2} + 2 x + 10 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

Solution steps

\int \frac{2 x + 1}{x ^{2} + 2 x + 10} d x

Complete the square

x^{2} + 2 x + 10 : (x + 1)^{2} + 9

= \int \frac{2 x + 1}{( x + 1 ) ^{2} + 9} d x

Apply u-substitution

: \int \frac{2 u - 1}{u ^{2} + 9} d u

= \int \frac{2 u - 1}{u ^{2} + 9} d u

Expand

\frac{2 u - 1}{u ^{2} + 9} : \frac{2 u}{u ^{2} + 9} - \frac{1}{u ^{2} + 9}

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} + 9} d u - \int \frac{1}{u ^{2} + 9} d u

\int \frac{2 u}{u ^{2} + 9} d u = ln u^{2} + 9

\int \frac{1}{u ^{2} + 9} d u = \frac{1}{3} arctan (\frac{u}{3})

= ln u^{2} + 9 - \frac{1}{3} arctan (\frac{u}{3})

Substitute back

u = x + 1

= ln (x + 1)^{2} + 9 - \frac{1}{3} arctan (\frac{x + 1}{3})

Simplify

ln (x + 1)^{2} + 9 - \frac{1}{3} arctan (\frac{x + 1}{3}) : ln x^{2} + 2 x + 10 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

= ln x^{2} + 2 x + 10 - \frac{1}{3} arctan (\frac{1}{3} (x + 1))

Add a constant to the solution

= ln x^{2} + 2 x + 10 - \frac{1}{3} arctan (\frac{1}{3} (x + 1)) + C

x \to 2 lim (x^{2} - 3 x + 1)

x \to 2 lim ([1 + f (x)]^{2})

\frac{\partial}{\partial x} (\frac{2 x}{z - y})

\int_{0}^{\frac{π}{3}} 10 sec^{2} (x) d x

\int_{- 7}^{7} 49 - x^{2} d x

What is the integral of (2x+1)/(x^2+2x+10) ?
The integral of (2x+1)/(x^2+2x+10) is ln|x^2+2x+10|-1/3 arctan(1/3 (x+1))+C