Question

evaluate the indefinite integral.
$$ int \frac { 4 } { ( t + 1 ) ^ { 4 } } d t $$
question help: video

Explanation:

Step1: Use substitution

Let \(u = t + 1\), then \(du=dt\).
The integral becomes \(\int\frac{4}{u^{4}}du\).

Step2: Rewrite the integrand

\(\int4u^{- 4}du\).

Step3: Integrate using power rule

\(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)).
For \(\int4u^{-4}du\), we have \(4\times\frac{u^{-4 + 1}}{-4 + 1}+C=4\times\frac{u^{-3}}{-3}+C=-\frac{4}{3u^{3}}+C\).

Step4: Substitute back \(u=t + 1\)

\(-\frac{4}{3(t + 1)^{3}}+C\).

Answer:

\(-\frac{4}{3(t + 1)^{3}}\)