Question

❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

We have to evaluate the given integral.

$= \displaystyle \sf \int x {(5 {x}^{2} + 3)}^{5} dx$

Let us assume that:

$\sf:\longmapsto u = 5{x}^{2} + 3$

$\sf:\longmapsto du = 10x \: dx$

$\sf:\longmapsto \dfrac{1}{10} du =x \: dx$

Therefore, the integral becomes:

$= \displaystyle \sf \int \frac{ {u}^{5} }{10} du$

$= \displaystyle \sf \dfrac{1}{10} \int {u}^{5} du$

$= \displaystyle \sf \dfrac{1}{10} \times \dfrac{ {u}^{5 + 1} }{5 + 1} + C$

$= \displaystyle \sf \dfrac{ {u}^{6} }{60} + C$

Substitute back u = 5x² + 3, we get:

$= \displaystyle \sf \dfrac{ {(5 {x}^{2} + 3 )}^{6} }{60} + C$

Therefore:

$\displaystyle \sf: \longmapsto \int x {(5 {x}^{2} + 3)}^{5} dx = \dfrac{ {(5 {x}^{2} + 3)}^{6} }{60} + C$

$\textsf{\large{\underline{Additional Information}:}}$

$\boxed{\begin{array}{c|c}\bf f(x)&\bf\displaystyle\int\rm f(x)\:dx\\ \\ \frac{\qquad\qquad}{}&\frac{\qquad\qquad}{}\\ \sf k&\sf kx+C\\ \\ \sf sin(x)&\sf-cos(x)+C\\ \\ \sf cos(x)&\sf sin(x)+C\\ \\ \sf{sec}^{2}(x)&\sf tan(x)+C\\ \\ \sf{cosec}^{2}(x)&\sf-cot(x)+C\\ \\ \sf sec(x)\ tan(x)&\sf sec(x)+C\\ \\ \sf cosec(x)\ cot(x)&\sf-cosec(x)+C\\ \\ \sf tan(x)&\sf log(sec(x))+C\\ \\ \sf\dfrac{1}{x}&\sf logx+C\\ \\ \sf{e}^{x}&\sf{e}^{x}+C\\ \\ \sf x^{n},n\neq-1&\sf\dfrac{x^{n+1}}{n+1}+C\end{array}}$

Answer 2

$\int{x \left(5 x^{2} + 3\right)^{5} d x}$

$Let \: u=5 x^{2} + 3$

$\color{red}Then \: du=\left(5 x^{2} + 3\right)^{\prime }dx = 10 x \: dx \: and \: we \: have \: that \: x \: dx = \frac{du}{10}$

The integral can be rewritten as

$\color{red}{\int{x \left(5 x^{2} + 3\right)^{5} d x}} = \color{red}{\int{\frac{u^{5}}{10} d u}}$

Apply the constant multiple rule

$\small{ \bold \red{\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du \: with \: c=\frac{1}{10} \: and \: f{\left(u \right)} = u^{5}:}}$

$\color{red}{\int{\frac{u^{5}}{10} d u}} = \color{red}{\left(\frac{\int{u^{5} d u}}{10}\right)}$

Apply the power rule

${ \bold \red{\int u^{n}\, du = \frac{u^{n + 1}}{n + 1} \left(n \neq -1 \right) with \: n=5:}}$

$\bold \red{\frac{\color{red}{\int{u^{5} d u}}}{10}=\frac{\color{red}{\frac{u^{1 + 5}}{1 + 5}}}{10}=\frac{\color{red}{\left(\frac{u^{6}}{6}\right)}}{10}}$

$\bold \red{Recall \: that \: u=5 x^{2} + 3}$

$\bold \red{\frac{\color{red}{u}^{6}}{60} = \frac{\color{red}{\left(5 x^{2} + 3\right)}^{6}}{60}}$

Therefore,

$\bold \red{\int{x \left(5 x^{2} + 3\right)^{5} d x} = \frac{\left(5 x^{2} + 3\right)^{6}}{60}}$

Add the constant of integration:

$\bold \red{\int{x \left(5 x^{2} + 3\right)^{5} d x} = \frac{\left(5 x^{2} + 3\right)^{6}}{60}+C}$

Answer:

$\bold \red{\int{x \left(5 x^{2} + 3\right)^{5} d x}=\frac{\left(5 x^{2} + 3\right)^{6}}{60}+C}$