find the indefinite integral and check the result by differentiating. \n\n∫x√x + 3dx\n\n∫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\neq - 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)