What is the integral of (3x)/(x^2+2x+5) ?

The integral of (3x)/(x^2+2x+5) is 3(1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1)))+C

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integral of (3x)/(x^2+2x+5)

\int \frac{3 x}{x ^{2} + 2 x + 5} d x

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

Solution steps

\int \frac{3 x}{x ^{2} + 2 x + 5} d x

Take the constant out:

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

= 3 \cdot \int \frac{x}{x ^{2} + 2 x + 5} d x

Complete the square

x^{2} + 2 x + 5 : (x + 1)^{2} + 4

= 3 \cdot \int \frac{x}{( x + 1 ) ^{2} + 4} d x

Apply u-substitution

= 3 \cdot \int \frac{u - 1}{u ^{2} + 4} d u

Expand

\frac{u - 1}{u ^{2} + 4} : \frac{u}{u ^{2} + 4} - \frac{1}{u ^{2} + 4}

Apply the Sum Rule:

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

= 3 (\int \frac{u}{u ^{2} + 4} d u - \int \frac{1}{u ^{2} + 4} d u)

\int \frac{u}{u ^{2} + 4} d u = \frac{1}{2} ln u^{2} + 4

\int \frac{1}{u ^{2} + 4} d u = \frac{1}{2} arctan (\frac{u}{2})

= 3 (\frac{1}{2} ln u^{2} + 4 - \frac{1}{2} arctan (\frac{u}{2}))

Substitute back

u = x + 1

= 3 (\frac{1}{2} ln (x + 1)^{2} + 4 - \frac{1}{2} arctan (\frac{x + 1}{2}))

Simplify

3 (\frac{1}{2} ln (x + 1)^{2} + 4 - \frac{1}{2} arctan (\frac{x + 1}{2})) : 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)))

= 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1)))

Add a constant to the solution

= 3 (\frac{1}{2} ln x^{2} + 2 x + 5 - \frac{1}{2} arctan (\frac{1}{2} (x + 1))) + C

What is the integral of (3x)/(x^2+2x+5) ?
The integral of (3x)/(x^2+2x+5) is 3(1/2 ln|x^2+2x+5|-1/2 arctan(1/2 (x+1)))+C