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

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

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

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

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

Solution steps

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

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

assuming

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

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Trigonometric Substitution

: \int \frac{1}{4 sec ( v )} d v

= \int \frac{1}{4 sec ( v )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{4} \cdot \int \frac{1}{sec ( v )} d v

Rewrite using trig identities

= \frac{1}{4} \cdot \int cos (v) d v

Use the common integral:

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

= \frac{1}{4} sin (v)

Substitute back

= \frac{1}{4} sin (arctan (\frac{1}{2} (x - 1)))

Simplify

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

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

Add a constant to the solution

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

\int \frac{1}{x ( x - 1 )} d x

x y^{^{'}} + 2 y = 2 e x

\frac{d}{d x} (sec^{2} (3 x))

y^{^{''}} - 2 \cdot y^{^{'}} = sin (x)

\int_{0}^{24} 9 + 3 x d x

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