What is the integral of 3/((z^2+4)(z+1)^2) ?

The integral of 3/((z^2+4)(z+1)^2) is 3(1/25 (-ln|z^2+4|-3/2 arctan(z/2))+2/25 ln|z+1|-1/(5(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 3/((z^2+4)(z+1)^2)

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

3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )}) + C

Solution steps

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

Take the constant out:

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

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

Take the partial fraction of

\frac{1}{( z ^{2} + 4 ) ( z + 1 ) ^{2}} : \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} + \frac{2}{25 ( z + 1 )} + \frac{1}{5 ( z + 1 ) ^{2}}

= 3 \cdot \int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} + \frac{2}{25 ( z + 1 )} + \frac{1}{5 ( z + 1 ) ^{2}} d z

Apply the Sum Rule:

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

= 3 (\int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} d z + \int \frac{2}{25 ( z + 1 )} d z + \int \frac{1}{5 ( z + 1 ) ^{2}} d z)

\int \frac{- 2 z - 3}{25 ( z ^{2} + 4 )} d z = \frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2}))

\int \frac{2}{25 ( z + 1 )} d z = \frac{2}{25} ln ∣ z + 1 ∣

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

= 3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )})

Add a constant to the solution

= 3 (\frac{1}{25} (- ln z^{2} + 4 - \frac{3}{2} arctan (\frac{z}{2})) + \frac{2}{25} ln ∣ z + 1 ∣ - \frac{1}{5 ( z + 1 )}) + C

\int cos (x) sin (2 x) d x

x \to - 1 lim (\frac{x ^{7} + 1}{x + 1})

\int (x^{6} + 1) e^{x^{7} + 7 x} d x

\int \frac{arcsin ( 2 x )}{1 - 4 x ^{2}} d x

f (x) = x + 5, at x = - 4

What is the integral of 3/((z^2+4)(z+1)^2) ?
The integral of 3/((z^2+4)(z+1)^2) is 3(1/25 (-ln|z^2+4|-3/2 arctan(z/2))+2/25 ln|z+1|-1/(5(z+1)))+C