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

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

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integral of (1+x)/(sqrt((x^2-2x-3)^3))

\int \frac{1 + x}{( x ^{2} - 2 x - 3 ) ^{3}} d x

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

Solution steps

\int \frac{1 + x}{( x ^{2} - 2 x - 3 ) ^{3}} d x

Complete the square

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

= \int \frac{1 + x}{( ( x - 1 ) ^{2} - 4 ) ^{3}} d x

Apply u-substitution

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

Expand

\frac{u + 2}{( u ^{2} - 4 ) ^{\frac{3}{2}}} : \frac{u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} + \frac{2}{( u ^{2} - 4 ) ^{\frac{3}{2}}}

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

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

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

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

= - \frac{1}{u ^{2} - 4} - \frac{u}{2 u ^{2} - 4}

Substitute back

u = x - 1

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

Simplify

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

= - \frac{1}{x ^{2} - 2 x - 3} - \frac{x - 1}{2 x ^{2} - 2 x - 3}

Add a constant to the solution

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

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