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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Find the value of integral ∫x³/(x+1) dx ?

 

Consider the integral

 \[\int \frac{x^{3}}{(x+1)}\; dx = \int \frac{x^{3}+1 -1}{(x+1)}\; dx\]

We have the algebraic identity: $\displaystyle x^{3}+1= (x+1)(x^{2}-x+1)$ .

Hence 

\[\begin{align*} \int \frac{x^{3}+1 -1}{(x+1)}\; dx&=\int \left (\frac{(x^{3}+1)}{x+1} - \frac{1}{(x+1)} \right )\; dx \\ &=\int \frac{x^{3}+1}{x+1}\; dx - \int\frac{1}{x+1} \; dx \\ &=\int \frac{(x+1)(x^{2}-x+1)}{x+1}\; dx -\int \frac{1}{x+1}\; dx\\ &=\int (x^{2}-x+1)\; dx -\int \frac{1}{x+1}\; dx \; \; (\mathbf{Cancelling \; x+1})\\ &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

And \[\begin{align*} \int \frac{x^{3}+1}{(x+1)}\; dx &= \frac{x^{3}}{3}-\frac{x^{2}}{2}+x- \log \left |x+1 \right |+C\\ \end{align*}\]

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Find the value of the integral $\displaystyle \int \frac{x+\sin x}{1+\cos x} dx$ =?

 

To find the value of the integral ,consider the following trigonometric formulas:

$\displaystyle \cos 2x= 2\cos ^{2} x-1 \Rightarrow \cos x=2\cos ^{2} \frac{x}{2} -1\Rightarrow 1+\cos x=2\cos ^{2} \frac{x}{2}\\ \sin 2x= 2 \sin x \cos x\Rightarrow \sin x= 2 \sin \frac{x}{2} \cos \frac{x}{2}$ 

Hence

  

\[\begin{align*} \int \frac{x+\sin x}{1+\cos x}\; dx&= \int \left (\frac{x}{1+\cos x}\; + \frac{\sin x}{1+\cos x}\; \right ) dx\\ &=\int \left (\frac{x}{2\cos^{2} \frac{x}{2}}\; + \frac{2\sin \frac{x}{2}\cos \frac{x}2{}}{2\cos^{2} \frac{x}{2}}\; \right ) dx \\ &= \int \left (\frac{1}{2} x \sec ^{2} \frac{x}{2}\; + \frac{\sin \frac{x}{2}}{\cos \frac{x}{2}}\; \right ) dx \\ &= \frac{1}{2}\int x\sec ^{2} \frac{x}{2} \; dx +\int \tan \frac{x}{2} dx\\ &=\frac{1}{2}\left (x\int \sec ^{2} \frac{x}{2} \; dx -\int \left (\frac{\mathrm{d} x}{\mathrm{d} x}\int \sec ^{2}\frac{x}{2}\; dx\right ) dx \right ) +\int \tan \frac{x}{2} dx \\ &= \frac{1}{2}\left (x\; \frac{\tan \frac{x}{2}}{1/2}-\int \frac{\tan \frac{x}{2}}{1/2}dx \right )+\int \tan \frac{x}{2}dx \\&=x\tan \frac{x}{2}+C-\int \tan \frac{x}{2}dx+\int \tan \frac{x}{2}dx\\ &=x\tan \frac{x}{2}+C \end{align*}\]

\[\therefore \int \frac{(x+\sin x)}{1+\cos x} \; dx= x \; \tan \frac{x}{2} +C\]

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If ∫x³ dx/(√(1+x²) = a (1+ x²)³/² +b√(1+x²)+ C, then a and b is given by:$\\ \mathbf{\;A)\; a=\frac{1}{3} , b=1.\; B)\; a=-\frac{1}{3}, b=1\; \\ C)\; a=-\frac{1}{3}, b=-1\; D)\; a=\frac{1}{3},b=-1} $

 

We have 

$\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx= a(1+x^{2})^{\frac{3}{2}}+b\sqrt{(1+x^{2})}+C \cdots (1)$ 

In-order  to find value of a and b , at first we have to evaluate the integral $\displaystyle \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx$ .

Now

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

Take 

$\displaystyle 1+x^{2}=u\Rightarrow 2x=\frac{\mathrm{d} u}{\mathrm{d} x}\mathbf{(Differentiating \; w.r.t \; x)}\\  \Rightarrow x \; dx=\frac{du}{2}$ 

Now by substituting $\displaystyle 1+x^{2}=u$  we will get

\[\begin{align*} \int \frac{x^{3}}{\sqrt{(1+x^{2})}} \; dx &= \int \frac{u-1}{\sqrt{u}} \frac{du}{2}\\ &= \frac{1}{2}\int \frac{u}{\sqrt{u}}- \frac{1}{\sqrt{u}} \; du\\ &=\frac{1}{2}[\int \frac{\sqrt{u}}{1}\; du-\int \frac{1}{\sqrt{u}}\; du] \\ &= \frac{1}{2}[\int u^{\frac{1}{2}}\; du-\int u^{-\frac{1}{2}}\; du]\\ &=\frac{1}{2}[\frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}-\frac{u^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}]+C \\ &=\frac{1}{2}[\frac{u^{3/2}}{3/2}-\frac{u^{1/2}}{1/2}] +C \\&= \frac{u^{3/2}}{3}-u^{1/2}+C \end{align*}\]

Now by providing value $\displaystyle u=1+x^{2}$  we will get:

$\displaystyle \int\frac{ x^{3}}{\sqrt{(1+x^{2})}} \; dx=\frac{1}{3}\left (1+x^{2} \right )^{\frac{3}{2}}-\left ( 1+x^{2} \right )^{\frac{1}{2}} +C \cdots(2)$ 

Comparing (1) and (2) we will get

$\displaystyle \mathbf{a=1/3\; and \; b=-1}$ 

Hence correct option :D

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$\displaystyle \int_{-\frac{\pi}{4}}^{\frac{\pi }{4}} \frac{dx}{1+\cos 2x}=\\ a) \;\; 1 \; \; b)\;\; 2\; \; c)\; \; 3\; \; d)\; \; 4$

 

Let $\displaystyle f(x)=\frac{1}{1+\cos 2x}\\$ 

By the properties of definite integral we have

$\mathbf{\int_{-a}^{a} f(x) dx} =\left\{\begin{matrix} 2\int_{0}^{a} f(x)dx \; \; if\; \textbf{f is even}, f(-x)=f(x)\\ 0 \;\; \; \; \; \; \; \; \;\; \; \; \; \; \; \: if \: \textbf{ f is odd},f(-x)=-f(x) \end{matrix}\right.$ 

Here $\displaystyle f(-x)=f(x)$ 

$\displaystyle f(-x)=\frac{1}{1+\cos (-2x)}=\frac{1}{1+\cos 2x}=f(x) \; (\because \cos (-x)=\cos x)$ 

Hence 

$\displaystyle f(x)=\frac{1}{1+\cos 2x}$   is an even function

\[\therefore \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx\]

And

\[\begin{align*} \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx &=2\int_{0}^{\frac{\pi }{4}} \frac{1}{1+2\cos^{2} x-1} \; dx \; \; (\because \cos 2x=2 \cos ^{2} x-1) \\ &= 2\int_{0}^{\frac{\pi }{4}} \frac{1}{2\cos^{2} x} \; dx\; \\ &= \int_{0}^{\frac{\pi }{4}} \frac{1}{\cos^{2} x} \; dx \; (by\; \; cancelling \; 2's)\\ &=\int_{0}^{\frac{\pi }{4}}\sec^{2} x\; dx \\ &= \left [ \tan x \right ]_{0}^{\frac{\pi }{4}}\\ &=\tan \frac{\pi }{4}-\tan 0 \\ &=1 \end{align*}\]

Hence $\displaystyle \int_{-\frac{\pi }{4}}^{\frac{\pi }{4}} \frac{1}{1+\cos 2x} \; dx =1$ 

And option a) 1 is the correct answer

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