What is the integral of x/(sqrt(4x^2+4x+5)) ?

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

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integral of x/(sqrt(4x^2+4x+5))

\int \frac{x}{4 x ^{2} + 4 x + 5} d x

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

Solution steps

\int \frac{x}{4 x ^{2} + 4 x + 5} d x

Complete the square

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

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

Apply u-substitution

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

Take the constant out:

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

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

Expand

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

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

Apply the Sum Rule:

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

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

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

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

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

Substitute back

u = x + \frac{1}{2}

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

Simplify

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

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

Add a constant to the solution

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

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