Question

find the indefinite integral and check the result by differentiating.

\\( \int x \sqrt { x + 4 } d x \\)

determine an appropriate substitution to simplify the integrand. choose the correct answer below.

\\( \bigcirc \\) a. \\( u = x - 4 \\)

\\( \bigcirc \\) b. \\( u = x + 4 \\)

\\( \bigcirc \\) c. \\( u = \sqrt { x } \\)

\\( \bigcirc \\) d. \\( u = \sqrt { x + 4 } \\)

find the indefinite integral.

\\( \int x \sqrt { x + 4 } d x = \square \\)

Explanation:

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

If \(u=x + 4\), then \(x=u - 4\) and \(dx=du\).
The integral \(\int x\sqrt{x + 4}dx\) becomes \(\int(u - 4)\sqrt{u}du=\int(u - 4)u^{\frac{1}{2}}du=\int(u^{\frac{3}{2}}-4u^{\frac{1}{2}})du\).

Step2: Integrate term - by - term

Using the power rule \(\int x^{n}dx=\frac{x^{n + 1}}{n+1}+C(n
eq - 1)\), we have:
\(\int(u^{\frac{3}{2}}-4u^{\frac{1}{2}})du=\int u^{\frac{3}{2}}du-4\int u^{\frac{1}{2}}du\)
\(=\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}-4\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)
\(=\frac{u^{\frac{5}{2}}}{\frac{5}{2}}-4\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C\)
\(=\frac{2}{5}u^{\frac{5}{2}}-\frac{8}{3}u^{\frac{3}{2}}+C\).

Step3: Substitute back \(u=x + 4\)

\(\frac{2}{5}(x + 4)^{\frac{5}{2}}-\frac{8}{3}(x + 4)^{\frac{3}{2}}+C\).

Step4: Check by differentiating

Let \(y=\frac{2}{5}(x + 4)^{\frac{5}{2}}-\frac{8}{3}(x + 4)^{\frac{3}{2}}+C\).
Using the chain rule \((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)\), where \(g(x)=x + 4\) and \(f(u)=\frac{2}{5}u^{\frac{5}{2}}-\frac{8}{3}u^{\frac{3}{2}}\) (\(u = x+4\)).
\(y^\prime=\frac{2}{5}\times\frac{5}{2}(x + 4)^{\frac{3}{2}}-\frac{8}{3}\times\frac{3}{2}(x + 4)^{\frac{1}{2}}\)
\(=(x + 4)^{\frac{3}{2}}-4(x + 4)^{\frac{1}{2}}\)
\(=(x + 4)^{\frac{1}{2}}[(x + 4)-4]\)
\(=x\sqrt{x + 4}\).

Answer:

\(\frac{2}{5}(x + 4)^{\frac{5}{2}}-\frac{8}{3}(x + 4)^{\frac{3}{2}}+C\)