What is the integral of (3x)/(x^2+10x+30) ?

The integral of (3x)/(x^2+10x+30) is 3(1/2 ln| 1/5 (x+5)^2+1|-sqrt(5)arctan(1/(sqrt(5))(x+5)))+C

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integral of (3x)/(x^2+10x+30)

\int \frac{3 x}{x ^{2} + 10 x + 30} d x

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

Solution steps

\int \frac{3 x}{x ^{2} + 10 x + 30} 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} + 10 x + 30} d x

Complete the square

x^{2} + 10 x + 30 : (x + 5)^{2} + 5

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

Apply u-substitution

: \int \frac{u - 5}{u ^{2} + 5} d u

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

Apply Integral Substitution

= 3 \cdot \int \frac{v - 5}{v ^{2} + 1} d v

Expand

\frac{v - 5}{v ^{2} + 1} : \frac{v}{v ^{2} + 1} - \frac{5}{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

= 3 (\int \frac{v}{v ^{2} + 1} d v - \int \frac{5}{v ^{2} + 1} d v)

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

\int \frac{5}{v ^{2} + 1} d v = 5 arctan (v)

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

Substitute back

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of (3x)/(x^2+10x+30) ?
The integral of (3x)/(x^2+10x+30) is 3(1/2 ln| 1/5 (x+5)^2+1|-sqrt(5)arctan(1/(sqrt(5))(x+5)))+C