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

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

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

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

Solution steps

\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{1}{2}}}

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

Add a constant to the solution

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

y^{^{'''}} - 5 y^{^{''}} - 22 y^{^{'}} + 56 y = 0

y = (x^{3} - 2) (x^{2} - 3 x + 1), (2, - 6)

\frac{d x}{d t} = 7 x

4 x + 3 x^{2}

(1 - x^{2})^{- \frac{1}{2}}

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