Question

find the indefinite integral and check the result by differentiating.

∫x√x + 3dx

∫x√x + 3dx =

Explanation:

Step1: Use substitution

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

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:
\(\int(u^{\frac{3}{2}}-3u^{\frac{1}{2}})du=\int u^{\frac{3}{2}}du-3\int u^{\frac{1}{2}}du\)
\(=\frac{u^{\frac{3}{2}+1}}{\frac{3}{2}+1}-3\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\)
\(=\frac{2}{5}u^{\frac{5}{2}}-2u^{\frac{3}{2}}+C\).

Step3: Substitute back \(u=x + 3\)

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

Step4: Simplify the expression

\(\frac{2}{5}(x + 3)^{\frac{3}{2}}(x + 3)-2(x + 3)^{\frac{3}{2}}+C\)
\(=(x + 3)^{\frac{3}{2}}(\frac{2(x + 3)}{5}-2)+C\)
\(=(x + 3)^{\frac{3}{2}}(\frac{2x+6 - 10}{5})+C\)
\(=\frac{2}{5}(x + 3)^{\frac{3}{2}}(x - 2)+C\).

Step5: Check by differentiation

Let \(y=\frac{2}{5}(x + 3)^{\frac{3}{2}}(x - 2)+C\)
Using the product rule \((uv)^\prime=u^\prime v+uv^\prime\), where \(u=\frac{2}{5}(x + 3)^{\frac{3}{2}}\), \(u^\prime=\frac{2}{5}\times\frac{3}{2}(x + 3)^{\frac{1}{2}}=\frac{3}{5}(x + 3)^{\frac{1}{2}}\) and \(v=x - 2\), \(v^\prime = 1\)
\(y^\prime=\frac{3}{5}(x + 3)^{\frac{1}{2}}(x - 2)+\frac{2}{5}(x + 3)^{\frac{3}{2}}\)
\(=(x + 3)^{\frac{1}{2}}[\frac{3(x - 2)}{5}+\frac{2(x + 3)}{5}]\)
\(=(x + 3)^{\frac{1}{2}}(\frac{3x-6 + 2x+6}{5})\)
\(=x\sqrt{x + 3}\)

Answer:

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