What is the integral of (4x)/(4x^2-12x+13) ?

The integral of (4x)/(4x^2-12x+13) is 2(1/4 ln|4x^2-12x+13|+3/4 arctan(x-3/2))+C

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integral of (4x)/(4x^2-12x+13)

\int \frac{4 x}{4 x ^{2} - 12 x + 13} d x

2 (\frac{1}{4} ln 4 x^{2} - 12 x + 13 + \frac{3}{4} arctan (x - \frac{3}{2})) + C

Solution steps

\int \frac{4 x}{4 x ^{2} - 12 x + 13} d x

Take the constant out:

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

= 4 \cdot \int \frac{x}{4 x ^{2} - 12 x + 13} d x

Complete the square

4 x^{2} - 12 x + 13 : 4 (x - \frac{3}{2})^{2} + 4

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

\frac{2 u + 3}{4 u ^{2} + 4} : \frac{2 u}{4 u ^{2} + 4} + \frac{3}{4 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

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

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

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

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

Substitute back

u = x - \frac{3}{2}

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

Simplify

4 \cdot \frac{1}{2} (\frac{1}{4} ln 4 (x - \frac{3}{2})^{2} + 4 + \frac{3}{4} arctan (x - \frac{3}{2})) : 2 (\frac{1}{4} ln 4 x^{2} - 12 x + 13 + \frac{3}{4} arctan (x - \frac{3}{2}))

= 2 (\frac{1}{4} ln 4 x^{2} - 12 x + 13 + \frac{3}{4} arctan (x - \frac{3}{2}))

Add a constant to the solution

= 2 (\frac{1}{4} ln 4 x^{2} - 12 x + 13 + \frac{3}{4} arctan (x - \frac{3}{2})) + C

What is the integral of (4x)/(4x^2-12x+13) ?
The integral of (4x)/(4x^2-12x+13) is 2(1/4 ln|4x^2-12x+13|+3/4 arctan(x-3/2))+C