What is the integral of 1/(sqrt((x^2-4x+5)^3)) ?

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

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integral of 1/(sqrt((x^2-4x+5)^3))

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

\frac{x - 2}{x ^{2} - 4 x + 5} + C

Solution steps

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

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

assuming

(x^{2} - 4 x + 5) \geq 0

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int cos (v) d v

= \int cos (v) d v

Use the common integral:

\int cos (v) d v = sin (v)

= sin (v)

Substitute back

= sin (arctan (x - 2))

Simplify

sin (arctan (x - 2)) : \frac{x - 2}{x ^{2} - 4 x + 5}

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

Add a constant to the solution

= \frac{x - 2}{x ^{2} - 4 x + 5} + C

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