math exercises\na) $intsqrt{5 + 2x}dx$\nb) $int x(3x^{2}-4)^{5}dx$\nc) $int\frac{ln^{2}x}{x}dx$\nd)…

math exercises\na) $intsqrt{5 + 2x}dx$\nb) $int x(3x^{2}-4)^{5}dx$\nc) $int\frac{ln^{2}x}{x}dx$\nd) $int\frac{3}{sqrt{(5 - 2x)^{3}}}dx$\ne) $int\frac{1}{x^{2}}cos\frac{1}{x}dx$\nf) $int\frac{5}{sqrt3{1 - 6x}}dx$\ng) $intsinleft(\frac{3x - 5}{2}\right)dx$\nh) $int x^{2}sqrt3{6 - x^{3}}dx$\nj)\nk)\nl)\nm)\nn)\no)\np)\nq)

math exercises\na) $intsqrt{5 + 2x}dx$\nb) $int x(3x^{2}-4)^{5}dx$\nc) $int\frac{ln^{2}x}{x}dx$\nd) $int\frac{3}{sqrt{(5 - 2x)^{3}}}dx$\ne) $int\frac{1}{x^{2}}cos\frac{1}{x}dx$\nf) $int\frac{5}{sqrt3{1 - 6x}}dx$\ng) $intsinleft(\frac{3x - 5}{2}\right)dx$\nh) $int x^{2}sqrt3{6 - x^{3}}dx$\nj)\nk)\nl)\nm)\nn)\no)\np)\nq)

Answer

Explanation:

Step1: Use substitution for $\int\sqrt{5 + 2x}dx$

Let $u = 5+2x$, then $du=2dx$ and $dx=\frac{1}{2}du$. The integral becomes $\frac{1}{2}\int u^{\frac{1}{2}}du$.

Step2: Integrate $u^{\frac{1}{2}}$

Using the power - rule for integration $\int x^n dx=\frac{x^{n + 1}}{n+1}+C$ ($n\neq - 1$), we have $\frac{1}{2}\times\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C=\frac{1}{2}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{3}u^{\frac{3}{2}}+C$.

Step3: Substitute back $u = 5 + 2x$

The result is $\frac{1}{3}(5 + 2x)^{\frac{3}{2}}+C$.

Answer:

$\frac{1}{3}(5 + 2x)^{\frac{3}{2}}+C$