The value of integral ∫e^x (1-x/1+x²)² dx=___?

In order to find the value of given integral \[\int e^{x}\left ( \frac{1-x}{1+x^{2}} \right )^{2} dx\].

For that consider property of integral: $\displaystyle \int e^{x}\left [ f(x)+{f}'(x) \right ] dx=e^{x} f(x)+C$ 

Hence

 \[\begin{align*} e^{x}\left ( \frac{1-x}{1+x^{2}} \right )^{2}&= e^{x}\frac{(1-x)^{2}}{(1+x^{2})^{2}}\\ &= e^{x}\left [ \frac{1-2x+x^{2}}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1+x^{2}-2x}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1+x^{2}}{(1+x^{2})^{2}}-\frac{2x}{(1+x^{2})^{2}} \right ]\\ &= e^{x}\left [ \frac{1}{(1+x^{2})}-\frac{2x}{(1+x^{2})^{2}} \right ]\\ \end{align*}\]

Here take $\displaystyle f(x)=\frac{1}{(1+x^{2})}\; \mathbf{and}\; {f}'(x)= \frac{-2x}{(1+x^{2})^{2}}$ and now by using property of integrals 

\[\begin{align*} \int e^{x}\left [ \frac{1-x }{1+x^{2}}\right ]^{2} dx&=\int e^{x}\left [ \frac{1}{(1+x^{2})}-\frac{2x}{(1+x^{2})^{2}} \right ]dx \\ &=\int e^{x}[f(x)+{f}'(x)]\; dx \\ &=e^{x} f(x) +C\\ &= e^{x} \frac{1}{1+x^{2}}+C \end{align*}\]

Hence $\displaystyle \begin{align*} \int e^{x}\left [ \frac{1-x }{1+x^{2}}\right ]^{2} dx &= \frac{e^{x}}{1+x^{2}}+C \end{align*}$ 

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The value of integral ∫x⁹/(4x²+1)⁶ dx=___?

 

In order to find value of integral \[\int \frac{x^{9}}{\left ( 4x^{2}+1 \right )^{6}} dx\]

we are going to use substitution method .

\[\mathbf{Put\; \; \; } 4x^{2}+1=u.\\ \mathbf{Differentiating \; w.r.t.x,\; \; } 2.4x=du\Rightarrow x.dx=\frac{du}{8}.\\ \mathbf{And\; \; } 4 x^{2}=u-1\Rightarrow x^{2}=\frac{u-1}{4}\]

Now we have substitute these values before that consider the integral,

\[\begin{align*} \int \frac{x^{9}}{\left ( 4x^{2}+1 \right )^{6}} dx &= \int \frac{(x^{2})^{4}.x}{(4x^{2}+1)^{6}} \; dx \\ &= \int \frac{(\frac{u-1}{4})^{4}}{u^{6}} \frac{du}{8} \\ &= \int \frac{({u-1})^{4}}{4^{4}u^{4}u^{2}} \frac{du}{8}\\ &= \frac{1}{4^{4}.8} \int \left [ \frac{u-1}{u} \right ]^{4}\frac{1}{u^{2}} \; du\\ &=\frac{1}{4^{5}.2} \int \left [1 -\frac{1}{u} \right ]^{4}\frac{1}{u^{2}} \; du \end{align*}\]

Again by substituting 1-1/u=t.

\[\mathbf{Differentiating\; w.r.t\; u:} \; \; -(-1 u^{-2})=\frac{1}{u^{2}}=\frac{dt}{du} \Rightarrow \frac{1}{u^{2}} du=dt\]

Hence

\[\begin{align*} \frac{1}{4^{5}.2} \int \left [1 -\frac{1}{u} \right ]^{4}\frac{1}{u^{2}} \; du &= \frac{1}{4^{5}.2} \int \left [t \right ]^{4} \; dt\\ &= \frac{1}{4^{5}.2} \left [\frac{t^{5}}{5} +C\right ] \; \\ &= \frac{t^{5}}{4^{5}.10} +C \end{align*}\]

Now for finding the value of integral we have to substitute values of t and u.

 \[\begin{align*} \therefore \int \frac{x^{9}}{\left ( 4x^{2}+1 \right )^{6}}\; dx&= \frac{t^{5}}{10.4^{5}} +C\\ &= \frac{[1-\frac{1}{u}]^{5}}{10.4^{5}} +C\; \; (\mathbf{by \; substuiting \; t=1-\frac{1}{u}})\\ &= \frac{1}{10.4^{5}}\left [ 1-\frac{1}{4x^{2}+1} \right ]^{5} +C\; \; (\mathbf{by \; substuiting \; u=\frac{1}{4x^{2}+1}})\\ &=\frac{1}{10.4^{5}}\left [ \frac{4x^{2}+1-1}{4x^{2}+1} \right ]^{5} +C\\ &= \frac{1}{10.4^{5}}\left [ \frac{4x^{2}}{4x^{2}+1} \right ]^{5} +C\\ &= \frac{1}{10.4^{5}}\left [ \frac{4}{4+\frac{1}{x^{2}}} \right ]^{5} +C\; (Dividing \; by\; x^{2})\\ &= \frac{4^{5}}{10.4^{5}}\left [ \frac{1}{4+\frac{1}{x^{2}}} \right ]^{5} +C\;(Cancelling \; \; 4^{5}) \\&= \frac{1}{10.}\left [ {4+\frac{1}{x^{2}}} \right ]^{-5} +C\; \end{align*}\]

\[\begin{align*} \therefore \int \frac{x^{9}}{\left ( 4x^{2}+1 \right )^{6}}\; dx &= \frac{1}{10.}\left [ {4+\frac{1}{x^{2}}} \right ]^{-5} +C\; \end{align*}\]

Hence option (D) is correct

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Find value of 'a' and 'b' , If $\displaystyle \int \frac{dx}{(x+2)(x^{2}+1)}\\$ =a log|1+x²| + b tan¯¹ x+ ⅕ log|x+2|+C ?

Given \[\int \frac{dx}{(x+2)(x^{2}+1)}=a \log |x^{2}+1| +b\tan ^{-1}x+\frac{1}{5}\log |x+2|+C\]

In order to find value of a and b  we have to evaluate the integral

$\displaystyle \int \frac{dx}{(x+2)(x^{2}+1)}$  at first.

For that we are using method of partial fractions.

 $\displaystyle \frac{1}{(x+2)(x^{2}+1)}= \frac{A}{x+2}+\frac{Bx+C}{x^2+1}\\\\ \Rightarrow 1=\left [\frac{A}{x+2}+\frac{Bx+C}{x^2+1} \right ](x+2)(x^{2}+1) \\\\ \Rightarrow 1=A(x^{2}+1)+(Bx+C)(x+2)$ 

Now we have to find values of A,B and C,for that

$\displaystyle \mathbf{Put\; x=-2}\\\Rightarrow 1=A((-2)^{2}+1)\Rightarrow 1=5A \; \mathbf{and}\; A=\frac{1}5{}\\ \\ \mathbf{By \; comparing \; coefficients\: of \; x^{2}}:0=A+B\Rightarrow B=-\frac{1}{5} \\\\ \mathbf{By \; comparing \; coefficients\: of \; x}:0=2B+C \Rightarrow C=-2B=\frac{2}{5} \\\\$ 

Now by substituting  values of A,B and C we will get

\[\frac{1}{(x+2)(x^{2}+1)}= \frac{\frac{1}{5}}{x+2}+\frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}\\\\\]

And

\[\begin{align*} \int \frac{dx}{(x+2)(x^{2}+1)}&= \int \frac{\frac{1}{5}}{x+2}\; dx+\int \frac{-\frac{1}{5}x+\frac{2}{5}}{x^2+1}\; dx \\ &= \frac{1}{5}\int \frac{dx}{x+2}-\int \left [\frac{1}{5} \frac{x}{x^{2}+1}+ \frac{2}{5}\frac{1}{x^{2}+1} \right ]dx\\ &= \frac{1}{5}\int \frac{dx}{x+2}-\frac{1}{5}\int \frac{x}{x^{2}+1} dx+ \frac{2}{5}\int \frac{1}{x^{2}+1} dx \end{align*}\]

Consider $\displaystyle \int \frac{x}{x^{2}+1}\; dx$ :

$\displaystyle \mathbf{Put \; x^{2}+1=u}\\ \mathbf{And \; differentiating \; w.r.t \; x .,2x\; dx=du\Rightarrow x\; dx=\frac{du}{2}} \\ \boldsymbol{\Rightarrow \int \frac{x}{x^{2}+1}\; dx=\int \frac{1}{u} \frac{du}{2}=\frac{1}{2}\log |u|+C }$ 

Hence

\[\begin{align*} \int \frac{dx}{(x+2)(x^{2}+1)} &=\frac{1}{5}\int \frac{dx}{x+2}-\frac{1}{5}\int \frac{x}{x^{2}+1} dx+ \frac{2}{5}\int \frac{1}{x^{2}+1} dx \\ &=\frac{1}{5}\log |x+2| -\frac{1}{5} \left [ \frac{1}{2} \log |x^{2}+1| \right ]+\frac{2}{5}\tan ^{-1}x+C\\ &=\frac{1}{5}\log |x+2| -\frac{1}{10} \left [ \log |x^{2}+1| \right ]+\frac{2}{5}\tan ^{-1}x+C \\ &=-\frac{1}{10} \log |x^{2}+1| +\frac{2}{5}\tan ^{-1}x+\frac{1}{5}\log |x+2|+C \end{align*}\\ \therefore \int \frac{dx}{(x+2)(x^{2}+1)}= -\frac{1}{10} \log |x^{2}+1| +\frac{2}{5}\tan ^{-1}x+\frac{1}{5}\log |x+2|+C\]

Hence \[a=-\frac{1}{10} \; \; \; \; \; \, \mathbf{and}\; \; \; \; \; b=\frac{2}{5}.\]

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