Question

find the following indefinite integral. \\(\int \frac{3t^2}{(5t^3 + 8)^2} dt = \square + c\\)

Explanation:

Step1: Choose substitution

Let \( u = 5t^{3}+8 \). Then, find the derivative of \( u \) with respect to \( t \):
\( \frac{du}{dt}=15t^{2} \), so \( du = 15t^{2}dt \), and we can solve for \( t^{2}dt \): \( t^{2}dt=\frac{du}{15} \).

Step2: Rewrite the integral

The integral is \( \int\frac{3t^{2}}{(5t^{3}+8)^{2}}dt \). Substitute \( u = 5t^{3}+8 \) and \( t^{2}dt=\frac{du}{15} \):
\( \int\frac{3\cdot\frac{du}{15}}{u^{2}}=\frac{3}{15}\int u^{-2}du \)
Simplify \( \frac{3}{15} \) to \( \frac{1}{5} \), so we have \( \frac{1}{5}\int u^{-2}du \).

Step3: Integrate using power rule

The power rule for integration is \( \int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\( n
eq - 1 \)). For \( n=-2 \):
\( \frac{1}{5}\cdot\frac{u^{-2 + 1}}{-2+1}+C=\frac{1}{5}\cdot\frac{u^{-1}}{-1}+C=-\frac{1}{5u}+C \)

Step4: Substitute back \( u \)

Substitute \( u = 5t^{3}+8 \) back into the expression:
\( -\frac{1}{5(5t^{3}+8)}+C \)

Answer:

\( -\frac{1}{5(5t^{3}+8)} \)