What is the integral of (-2z-3)/(25(z^2+4)) ?

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

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integral of (-2z-3)/(25(z^2+4))

\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})) + C

Solution steps

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

Take the constant out:

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

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

Expand

\frac{- 2 z - 3}{z ^{2} + 4} : - \frac{2 z}{z ^{2} + 4} - \frac{3}{z ^{2} + 4}

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

\int \frac{2 z}{z ^{2} + 4} d z = ln z^{2} + 4

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

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

Add a constant to the solution

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

y^{^{'}} + 2 y = 5 y^{3}

y^{^{''}} - 2 y^{^{'}} + y = 2 x e^{2 x} + 6 e^{x}

e^{s i n (x)}

\int (x^{7} - 3 x^{4} + x) d x

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

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