What is the integral of 1/((z^2-2z+5)^2) ?

The integral of 1/((z^2-2z+5)^2) is 1/16 (arctan(1/2 (z-1))+1/2 sin(2arctan(1/2 (z-1))))+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/((z^2-2z+5)^2)

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

\frac{1}{16} (arctan (\frac{1}{2} (z - 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (z - 1)))) + C

Solution steps

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

Complete the square

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

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

Apply u-substitution

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

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

Apply Integral Substitution

= \int \frac{1}{8 ( v ^{2} + 1 ) ^{2}} d v

Take the constant out:

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

= \frac{1}{8} \cdot \int \frac{1}{( v ^{2} + 1 ) ^{2}} d v

Apply Trigonometric Substitution

: \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

= \frac{1}{8} \cdot \int \frac{1}{sec ^{4} ( w ) cos ^{2} ( w )} d w

Rewrite using trig identities

= \frac{1}{8} \cdot \int \frac{1 + cos ( 2 w )}{2} d w

Take the constant out:

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

= \frac{1}{8} \cdot \frac{1}{2} \cdot \int 1 + cos (2 w) d w

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{8} \cdot \frac{1}{2} (\int 1 d w + \int cos (2 w) d w)

\int 1 d w = w

\int cos (2 w) d w = \frac{1}{2} sin (2 w)

= \frac{1}{8} \cdot \frac{1}{2} (w + \frac{1}{2} sin (2 w))

Substitute back

= \frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{z - 1}{2}) + \frac{1}{2} sin (2 arctan (\frac{z - 1}{2})))

Simplify

\frac{1}{8} \cdot \frac{1}{2} (arctan (\frac{z - 1}{2}) + \frac{1}{2} sin (2 arctan (\frac{z - 1}{2}))) : \frac{1}{16} (arctan (\frac{1}{2} (z - 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (z - 1))))

= \frac{1}{16} (arctan (\frac{1}{2} (z - 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (z - 1))))

Add a constant to the solution

= \frac{1}{16} (arctan (\frac{1}{2} (z - 1)) + \frac{1}{2} sin (2 arctan (\frac{1}{2} (z - 1)))) + C

\int \frac{4 x}{16 - x ^{2}} d x

\int 3 sec^{2} (3 x) tan (3 x) d x

\int \frac{( arcsin ( ln ( x ) ) )}{x} d x

\frac{1}{x ^{4}}

\int 2 sin (x) - e^{x} d x

What is the integral of 1/((z^2-2z+5)^2) ?
The integral of 1/((z^2-2z+5)^2) is 1/16 (arctan(1/2 (z-1))+1/2 sin(2arctan(1/2 (z-1))))+C