What is the integral of (10x)/(4x^2+8x+9) ?

The integral of (10x)/(4x^2+8x+9) is 5/4 ln| 4/5 (x+1)^2+1|-sqrt(5)arctan(2/(sqrt(5))(x+1))+C

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integral of (10x)/(4x^2+8x+9)

\int \frac{10 x}{4 x ^{2} + 8 x + 9} d x

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

Solution steps

\int \frac{10 x}{4 x ^{2} + 8 x + 9} d x

Take the constant out:

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

= 10 \cdot \int \frac{x}{4 x ^{2} + 8 x + 9} d x

Complete the square

4 x^{2} + 8 x + 9 : 4 (x + 1)^{2} + 5

= 10 \cdot \int \frac{x}{4 ( x + 1 ) ^{2} + 5} d x

Apply u-substitution

= 10 \cdot \int \frac{u - 1}{4 u ^{2} + 5} d u

Apply Integral Substitution

= 10 \cdot \int \frac{5 v - 2}{4 5 ( v ^{2} + 1 )} d v

Take the constant out:

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

= 10 \cdot \frac{1}{4 5} \cdot \int \frac{5 v - 2}{v ^{2} + 1} d v

Expand

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

= 10 \cdot \frac{1}{4 5} (\int \frac{5 v}{v ^{2} + 1} d v - \int \frac{2}{v ^{2} + 1} d v)

\int \frac{5 v}{v ^{2} + 1} d v = \frac{5}{2} ln v^{2} + 1

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

= 10 \cdot \frac{1}{4 5} (\frac{5}{2} ln v^{2} + 1 - 2 arctan (v))

Substitute back

= 10 \cdot \frac{1}{4 5} (\frac{5}{2} ln (\frac{2}{5} (x + 1))^{2} + 1 - 2 arctan (\frac{2}{5} (x + 1)))

Simplify

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

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

Add a constant to the solution

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

What is the integral of (10x)/(4x^2+8x+9) ?
The integral of (10x)/(4x^2+8x+9) is 5/4 ln| 4/5 (x+1)^2+1|-sqrt(5)arctan(2/(sqrt(5))(x+1))+C