calculate the indefinite integral: $intleft(6x^{2}+3sqrt{x}+\frac{1}{x^{3}}\right)dx$.\n$-\frac{1}{2x^{2}}+2x…

calculate the indefinite integral: $intleft(6x^{2}+3sqrt{x}+\frac{1}{x^{3}}\right)dx$.\n$-\frac{1}{2x^{2}}+2x^{3}+2x^{3/2}+c$\n$12x+\frac{3}{2sqrt{x}}-\frac{3}{x^{4}}+c$\n$-\frac{1}{x}+2x^{3}+2x^{3/2}+c$\n$-\frac{1}{2x^{2}}+3x^{2}-2x^{3/2}+c$\n$-\frac{1}{2x^{2}}+2x^{3}-2x^{3/2}+c$\nnone of the above.

calculate the indefinite integral: $intleft(6x^{2}+3sqrt{x}+\frac{1}{x^{3}}\right)dx$.\n$-\frac{1}{2x^{2}}+2x^{3}+2x^{3/2}+c$\n$12x+\frac{3}{2sqrt{x}}-\frac{3}{x^{4}}+c$\n$-\frac{1}{x}+2x^{3}+2x^{3/2}+c$\n$-\frac{1}{2x^{2}}+3x^{2}-2x^{3/2}+c$\n$-\frac{1}{2x^{2}}+2x^{3}-2x^{3/2}+c$\nnone of the above.

Answer

Explanation:

Step1: Integrate each term separately

Use the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$). For the first term $\int6x^{2}dx$: Let $n = 2$, then $\int6x^{2}dx=6\times\frac{x^{2 + 1}}{2+1}=2x^{3}$.

Step2: Integrate the second term

For the second term $\int3\sqrt{x}dx=\int3x^{\frac{1}{2}}dx$. Let $n=\frac{1}{2}$, then $\int3x^{\frac{1}{2}}dx=3\times\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}=3\times\frac{x^{\frac{3}{2}}}{\frac{3}{2}} = 2x^{\frac{3}{2}}$.

Step3: Integrate the third term

For the third term $\int\frac{1}{x^{3}}dx=\int x^{-3}dx$. Let $n=-3$, then $\int x^{-3}dx=\frac{x^{-3 + 1}}{-3+1}=\frac{x^{-2}}{-2}=-\frac{1}{2x^{2}}$.

Step4: Combine the results

$\int(6x^{2}+3\sqrt{x}+\frac{1}{x^{3}})dx=-\frac{1}{2x^{2}}+2x^{3}+2x^{\frac{3}{2}}+C$.

Answer:

$-\frac{1}{2x^{2}}+2x^{3}+2x^{\frac{3}{2}}+C$