What is the integral of (6z)/((5-3z^2)^2) ?

The integral of (6z)/((5-3z^2)^2) is 1/(5-3z^2)+C

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integral of (6z)/((5-3z^2)^2)

\int \frac{6 z}{( 5 - 3 z ^{2} ) ^{2}} d z

\frac{1}{5 - 3 z ^{2}} + C

Solution steps

\int \frac{6 z}{( 5 - 3 z ^{2} ) ^{2}} d z

Take the constant out:

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

= 6 \cdot \int \frac{z}{( 5 - 3 z ^{2} ) ^{2}} d z

Apply u-substitution

= 6 \cdot \int - \frac{1}{6 u ^{2}} d u

Take the constant out:

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

= 6 (- \frac{1}{6} \cdot \int \frac{1}{u ^{2}} d u)

Apply the Power Rule

= 6 (- \frac{1}{6} (- \frac{1}{u}))

Substitute back

u = 5 - 3 z^{2}

= 6 (- \frac{1}{6} (- \frac{1}{5 - 3 z ^{2}}))

Simplify

6 (- \frac{1}{6} (- \frac{1}{5 - 3 z ^{2}})) : \frac{1}{5 - 3 z ^{2}}

= \frac{1}{5 - 3 z ^{2}}

Add a constant to the solution

= \frac{1}{5 - 3 z ^{2}} + C

What is the integral of (6z)/((5-3z^2)^2) ?
The integral of (6z)/((5-3z^2)^2) is 1/(5-3z^2)+C