What is the integral of (2x-3)/((x^2+2x-3)^{3/2)} ?

The integral of (2x-3)/((x^2+2x-3)^{3/2)} is -2/((x^2+2x-3)^{1/2)}+(5(x+1))/(4(x^2+2x-3)^{1/2)}+C

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integral of (2x-3)/((x^2+2x-3)^{3/2)}

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

- \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

: \int \frac{2 u - 5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

= \int \frac{2 u - 5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

Expand

\frac{2 u - 5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} : \frac{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} - \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}}

= \int \frac{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} - \frac{5}{( 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{2 u}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u - \int \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u

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

\int \frac{5}{( u ^{2} - 4 ) ^{\frac{3}{2}}} d u = - \frac{5 u}{4 ( u ^{2} - 4 ) ^{\frac{1}{2}}}

= - \frac{2}{( u ^{2} - 4 ) ^{\frac{1}{2}}} - (- \frac{5 u}{4 ( u ^{2} - 4 ) ^{\frac{1}{2}}})

Substitute back

u = x + 1

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

Simplify

- \frac{2}{( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}} - - \frac{5 ( x + 1 )}{4 ( ( x + 1 ) ^{2} - 4 ) ^{\frac{1}{2}}} : - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}}

= - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}}

Add a constant to the solution

= - \frac{2}{( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + \frac{5 ( x + 1 )}{4 ( x ^{2} + 2 x - 3 ) ^{\frac{1}{2}}} + C

2^{t}

\int \frac{8 x - 2}{x ^{2} - 1} d x

\int e^{9 x} d x

x \to 2 lim (x + 6)

\frac{d}{d t} (x ln (t))

What is the integral of (2x-3)/((x^2+2x-3)^{3/2)} ?
The integral of (2x-3)/((x^2+2x-3)^{3/2)} is -2/((x^2+2x-3)^{1/2)}+(5(x+1))/(4(x^2+2x-3)^{1/2)}+C