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

The integral of (2x)/(sqrt(x^2-2x-3)) is 2(sqrt(x^2-2x-3)+ln(1/2 |sqrt(x^2-2x-3)+x-1|))+C

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integral of (2x)/(sqrt(x^2-2x-3))

\int \frac{2 x}{x ^{2} - 2 x - 3} d x

2 (x^{2} - 2 x - 3 + ln (\frac{1}{2} x^{2} - 2 x - 3 + x - 1)) + C

Solution steps

\int \frac{2 x}{x ^{2} - 2 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} - 2 x - 3} d x

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

\frac{d y}{d x} = e^{4 x}

x^{2} e^{- x^{2}}

x + 3

g (t) = 6 t^{3} cos (t)

f (x) = x y

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