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

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

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

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

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

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = x - 1

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

Simplify

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

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

Add a constant to the solution

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

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