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

x \to 2 lim ([- \frac{x}{2}] + x^{2})

\frac{\partial}{\partial y} (3 sin (x) sin (y))

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

e^{x}, 3

y = 5 x, at x = 4, x = 9

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