Question

evaluate the integral
int \frac { d x } { ( 4 x + 3 ) ^ { 4 } },
by making the appropriate substitution: ( u = )
int \frac { d x } { ( 4 x + 3 ) ^ { 4 } } =
question help: video

Explanation:

Step1: Substitute \(u = 4x+3\)

Differentiate \(u\) with respect to \(x\): \(du=4dx\), so \(dx=\frac{1}{4}du\).

Step2: Rewrite the integral

The integral \(\int\frac{dx}{(4x + 3)^{4}}\) becomes \(\int\frac{1}{u^{4}}\cdot\frac{1}{4}du=\frac{1}{4}\int u^{- 4}du\).

Step3: Integrate using the power rule

The power rule for integration is \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C\) (\(n
eq - 1\)). For \(n=-4\), \(\frac{1}{4}\int u^{-4}du=\frac{1}{4}\cdot\frac{u^{-4 + 1}}{-4+1}+C=\frac{1}{4}\cdot\frac{u^{-3}}{-3}+C=-\frac{1}{12u^{3}}+C\).

Step4: Substitute back \(u = 4x+3\)

We get \(-\frac{1}{12(4x + 3)^{3}}+C\).

Answer:

\(u = 4x+3\); \(\int\frac{dx}{(4x + 3)^{4}}=-\frac{1}{12(4x + 3)^{3}}+C\)