Question

Find the integration of the following function

$\frac{x - 1}{(x + 1 )\sqrt{ {x}^{3} + {x}^{2} + x } }$
​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle\int\rm \frac{(x - 1)}{(x + 1) \sqrt{( {x}^{3} + {x}^{2} + x)} } \: dx$

can be further rewritten as

$\rm \:  =  \: \:\displaystyle\int\rm \frac{(x - 1)(x + 1)}{(x + 1)^{2} \sqrt{( {x}^{3} + {x}^{2} + x)} } \: dx$

$\rm \:  =  \: \:\displaystyle\int\rm \frac{ {x}^{2} - 1}{(x^{2} + 1 + 2x) \sqrt{( {x}^{3} + {x}^{2} + x)} } \: dx$

$\rm \:  =  \: \:\displaystyle\int\rm \frac{ {x}^{2}\bigg(1 - \dfrac{1}{ {x}^{2} } \bigg)}{x\bigg(x+\dfrac{1}{x} + 2\bigg)x \sqrt{{x}+ 1+\dfrac{1}{ {x}} }} \: dx$

$\rm \:  =  \: \:\displaystyle\int\rm \frac{\bigg(1 - \dfrac{1}{ {x}^{2} } \bigg)}{\bigg(x+\dfrac{1}{x} + 2\bigg) \sqrt{{x}+ 1+\dfrac{1}{ {x}} }} \: dx$

$\rm \:  =  \: \:\displaystyle\int\rm \frac{\bigg(1 - \dfrac{1}{ {x}^{2} } \bigg)}{\bigg(x+\dfrac{1}{x} + 1 + 1\bigg) \sqrt{{x}+ 1+\dfrac{1}{ {x}} }} \: dx$

Now, to evaluate this integral further, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\: \sqrt{x + 1 + \dfrac{1}{x} } = y$

$\rm :\longmapsto\:x + 1 + \dfrac{1}{x} = {y}^{2}$

$\rm :\longmapsto\:\bigg(1 - \dfrac{1}{ {x}^{2} }\bigg)dx = 2ydy$

So, on substituting the values, we get

$\rm \:  =  \: \displaystyle\int\rm \frac{2y \: dy}{( {y}^{2} + 1)y }$

$\rm \:  =  \: 2\displaystyle\int\rm \frac{\: dy}{{y}^{2} + 1}$

$\rm \:  =  \: 2 {tan}^{ - 1}y + c$

$\rm \:  =  \: 2 {tan}^{ - 1} \sqrt{x + 1 + \dfrac{1}{x} } + c$

$\rm \:  =  \: 2 {tan}^{ - 1} \sqrt{ \dfrac{ {x}^{2} + x + 1}{x} } + c$

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LEARN MORE

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$

Answer 2

Question:-

Find the integration of the following function $\sf\large\frac{x - 1}{(x + 1 )\sqrt{ {x}^{3} + {x}^{2} + x } }.$

Given:-

$\sf \large\frac{x - 1}{(x + 1 )\sqrt{ {x}^{3} + {x}^{2} + x } } .$

To Find:-

• The integration of the following function.

Solution:-

[Refer to the above attachment]

Answer:-

[Refer to the above attachment]

Hope you have satisfied. ⚘