What is the integral of (x+3)/(x^2+2x+10) ?

The integral of (x+3)/(x^2+2x+10) is 1/2 ln|((x+1)^2)/9+1|+2/3 arctan(1/3 (x+1))+C

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integral of (x+3)/(x^2+2x+10)

\int \frac{x + 3}{x ^{2} + 2 x + 10} d x

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

Solution steps

\int \frac{x + 3}{x ^{2} + 2 x + 10} d x

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{3 v + 2}{3 ( v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{3 v + 2}{v ^{2} + 1} d v

Expand

\frac{3 v + 2}{v ^{2} + 1} : \frac{3 v}{v ^{2} + 1} + \frac{2}{v ^{2} + 1}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{3} (\int \frac{3 v}{v ^{2} + 1} d v + \int \frac{2}{v ^{2} + 1} d v)

\int \frac{3 v}{v ^{2} + 1} d v = \frac{3}{2} ln v^{2} + 1

\int \frac{2}{v ^{2} + 1} d v = 2 arctan (v)

= \frac{1}{3} (\frac{3}{2} ln v^{2} + 1 + 2 arctan (v))

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (x+3)/(x^2+2x+10) ?
The integral of (x+3)/(x^2+2x+10) is 1/2 ln|((x+1)^2)/9+1|+2/3 arctan(1/3 (x+1))+C