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

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

10 (\frac{1}{8} ln 4 x^{2} - 8 x + 9 + \frac{1}{2 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

Expand

\frac{u + 1}{4 u ^{2} + 5} : \frac{u}{4 u ^{2} + 5} + \frac{1}{4 u ^{2} + 5}

Apply the Sum Rule:

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

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

\int \frac{u}{4 u ^{2} + 5} d u = \frac{1}{8} ln 4 u^{2} + 5

\int \frac{1}{4 u ^{2} + 5} d u = \frac{1}{2 5} arctan (\frac{2}{5} u)

= 10 (\frac{1}{8} ln 4 u^{2} + 5 + \frac{1}{2 5} arctan (\frac{2}{5} u))

Substitute back

u = x - 1

= 10 (\frac{1}{8} ln 4 (x - 1)^{2} + 5 + \frac{1}{2 5} arctan (\frac{2}{5} (x - 1)))

Simplify

10 (\frac{1}{8} ln 4 (x - 1)^{2} + 5 + \frac{1}{2 5} arctan (\frac{2}{5} (x - 1))) : 10 (\frac{1}{8} ln 4 x^{2} - 8 x + 9 + \frac{1}{2 5} arctan (\frac{2}{5} (x - 1)))

= 10 (\frac{1}{8} ln 4 x^{2} - 8 x + 9 + \frac{1}{2 5} arctan (\frac{2}{5} (x - 1)))

Add a constant to the solution

= 10 (\frac{1}{8} ln 4 x^{2} - 8 x + 9 + \frac{1}{2 5} arctan (\frac{2}{5} (x - 1))) + C

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