What is the integral of (2x)/(sqrt(x^2+4x+3)) ?

The integral of (2x)/(sqrt(x^2+4x+3)) is 2(sqrt(x^2+4x+3)-2ln|sqrt(x^2+4x+3)+x+2|)+C

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integral of (2x)/(sqrt(x^2+4x+3))

\int \frac{2 x}{x ^{2} + 4 x + 3} d x

2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2) + C

Solution steps

\int \frac{2 x}{x ^{2} + 4 x + 3} d x

Take the constant out:

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

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

Complete the square

x^{2} + 4 x + 3 : (x + 2)^{2} - 1

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

Apply u-substitution

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

Expand

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

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

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

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

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

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

Substitute back

u = x + 2

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

Simplify

2 ((x + 2)^{2} - 1 - 2 ln (x + 2)^{2} - 1 + x + 2) : 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2)

= 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2)

Add a constant to the solution

= 2 (x^{2} + 4 x + 3 - 2 ln x^{2} + 4 x + 3 + x + 2) + C

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