Question

Evaluate the following indefinite integral. Express Your answer in positive exponent

Answer 1

1.  Let,

$\displaystyle\longrightarrow\sf{u=3x+5}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{3}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int(3x+5)^7\ dx=\dfrac{1}{3}\int u^7\ du}$

$\displaystyle\longrightarrow\sf{\int(3x+5)^7\ dx=\dfrac{1}{3}\cdot\dfrac{u^8}{8}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int(3x+5)^7\ dx=\dfrac{(3x+5)^8}{24}+c}}}$

2.  Let,

$\displaystyle\longrightarrow\sf{u=6x+9}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{6}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt{6x+9}\ dx=\dfrac{1}{6}\int u^{\frac{1}{2}}\ du}$

$\displaystyle\longrightarrow\sf{\int\sqrt{6x+9}\ dx=\dfrac{1}{6}\cdot\dfrac{u^{\frac{3}{2}}}{\left(\dfrac{3}{2}\right)}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\sqrt{6x+9}\ dx=\dfrac{(6x+9)^{\frac{3}{2}}}{9}+c}}}$

3.  Let,

$\displaystyle\longrightarrow\sf{u=5x-1}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{5}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt[3]{\sf{5x-1}}\ dx=\dfrac{1}{5}\int u^{\frac{1}{3}}\ du}$

$\displaystyle\longrightarrow\sf{\int\sqrt[3]{\sf{5x-1}}\ dx=\dfrac{1}{5}\cdot\dfrac{u^{\frac{4}{3}}}{\left(\dfrac{4}{3}\right)}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\sqrt[3]{\sf{5x-1}}\ dx=\dfrac{3(5x-1)^{\frac{4}{3}}}{20}+c}}}$

4.  Let,

$\displaystyle\longrightarrow\sf{u=3x-5}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{3}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int3(3x-5)^4\ dx=3\cdot\dfrac{1}{3}\int u^4\ du}$

$\displaystyle\longrightarrow\sf{\int3(3x-5)^4\ dx=\dfrac{u^5}{5}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int3(3x-5)^4\ dx=\dfrac{(3x-5)^5}{5}+c}}}$

5.  Let,

$\displaystyle\longrightarrow\sf{u=3x+2}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{3}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int\sqrt[4]{\sf{(3x+2)^5}}\ dx=\dfrac{1}{3}\int u^{\frac{5}{4}}\ du}$

$\displaystyle\longrightarrow\sf{\int\sqrt[4]{\sf{(3x+2)^5}}\ dx=\dfrac{1}{3}\cdot\dfrac{u^{\frac{9}{4}}}{\left(\dfrac{9}{4}\right)}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\sqrt[4]{\sf{(3x+2)^5}}\ dx=\dfrac{4(3x+2)^{\frac{9}{4}}}{27}+c}}}$

7. (or 6.?)  Let,

$\displaystyle\longrightarrow\sf{u=2x-5}$

$\displaystyle\longrightarrow\sf{dx=\dfrac{1}{2}\ du}$

Then,

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{\sqrt{2x-5}}\ dx=\dfrac{1}{2}\int\dfrac{1}{u^{\frac{1}{2}}}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{\sqrt{2x-5}}\ dx=\dfrac{1}{2}\int u^{-\frac{1}{2}}\ du}$

$\displaystyle\longrightarrow\sf{\int\dfrac{1}{\sqrt{2x-5}}\ dx=\dfrac{1}{2}\cdot\dfrac{u^{\frac{1}{2}}}{\left(\dfrac{1}{2}\right)}+c}$

$\displaystyle\longrightarrow\sf{\underline{\underline{\int\dfrac{1}{\sqrt{2x-5}}\ dx=(2x-5)^{\frac{1}{2}}+c}}}$