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

The integral of (x-3)/((5-4x-x^2)^{3/2)} is 1/((-x^2-4x+5)^{1/2)}-(5(x+2))/(9(-x^2-4x+5)^{1/2)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (x-3)/((5-4x-x^2)^{3/2)}

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

\frac{1}{( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} + C

Solution steps

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

Complete the square

5 - 4 x - x^{2} : - (x + 2)^{2} + 9

= \int \frac{x - 3}{( - ( x + 2 ) ^{2} + 9 ) ^{\frac{3}{2}}} d x

Apply u-substitution

: \int \frac{u - 5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

= \int \frac{u - 5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

Expand

\frac{u - 5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} : \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} - \frac{5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}}

= \int \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} - \frac{5}{( - u ^{2} + 9 ) ^{\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} + 9 ) ^{\frac{3}{2}}} d u - \int \frac{5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u

\int \frac{u}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u = \frac{1}{( - u ^{2} + 9 ) ^{\frac{1}{2}}}

\int \frac{5}{( - u ^{2} + 9 ) ^{\frac{3}{2}}} d u = \frac{5 u}{9 ( 9 - u ^{2} ) ^{\frac{1}{2}}}

= \frac{1}{( - u ^{2} + 9 ) ^{\frac{1}{2}}} - \frac{5 u}{9 ( 9 - u ^{2} ) ^{\frac{1}{2}}}

Substitute back

u = x + 2

= \frac{1}{( - ( x + 2 ) ^{2} + 9 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( 9 - ( x + 2 ) ^{2} ) ^{\frac{1}{2}}}

Simplify

\frac{1}{( - ( x + 2 ) ^{2} + 9 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( 9 - ( x + 2 ) ^{2} ) ^{\frac{1}{2}}} : \frac{1}{( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}}

= \frac{1}{( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}}

Add a constant to the solution

= \frac{1}{( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} - \frac{5 ( x + 2 )}{9 ( - x ^{2} - 4 x + 5 ) ^{\frac{1}{2}}} + C

\frac{d}{d x} (0.8 x^{1.6})

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

\int_{2}^{4} \frac{5}{x} d x

k = 0 \sum \infty (- \frac{2}{9})^{k}

\frac{\partial}{\partial x} (\frac{y}{x} + ln (y))

What is the integral of (x-3)/((5-4x-x^2)^{3/2)} ?
The integral of (x-3)/((5-4x-x^2)^{3/2)} is 1/((-x^2-4x+5)^{1/2)}-(5(x+2))/(9(-x^2-4x+5)^{1/2)}+C