Explanation
The value of $$\displaystyle\int {\dfrac{{\ln n\left( {1 - \left({\dfrac{1}{x}} \right)} \right)dx}}{{x\left( {x - 1} \right)}}} $$ is
Consider, $$\displaystyle I=\int {{{ln\left( {1 - {1 \over x}} \right)} \over {x\left( {x - 1} \right)}}} dx$$
$$\displaystyle I = \int {{{\ln\left( {1 - {1 \over x}} \right)} \over {{x^2}\left( {1 - {1 \over x}} \right)}}} dx$$
Put $$\ln\left( {1 - {\dfrac 1 x}} \right) = t$$ $$\Rightarrow$$ $$\displaystyle {1 \over {\left( {1 - {\dfrac 1 x}} \right)}} \times {\dfrac {1} {{x^2}}}dx = dt$$
$$\displaystyle I= \int {tdt} $$
$$\displaystyle I= {{{t^2}} \over 2} + c$$
$$\displaystyle I= {1 \over 2}{\left[ {ln\left( {1 - {1 \over x}} \right)} \right]^2} + c$$
Consider the given equation.
$$I=\int{{{e}^{{{\tan }^{-1}}x}}\left( \dfrac{1+x+{{x}^{2}}}{1+{{x}^{2}}} \right)}dx$$
$$I=\int{\left( \dfrac{{{e}^{{{\tan }^{-1}}x}}+x{{e}^{{{\tan }^{-1}}x}}+{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} \right)}dx$$ ……… (1)
Let $$t=x{{e}^{{{\tan }^{-1}}x}}$$
$$ t=x{{e}^{{{\tan }^{-1}}x}} $$
$$ \dfrac{dt}{dx}={{e}^{{{\tan }^{-1}}x}}\times 1+x\left( {{e}^{{{\tan }^{-1}}x}} \right)\times \dfrac{1}{1+{{x}^{2}}} $$
$$ \dfrac{dt}{dx}={{e}^{{{\tan }^{-1}}x}}+\dfrac{x{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} $$
$$ dt=\left( \dfrac{{{e}^{{{\tan }^{-1}}x}}+{{x}^{2}}{{e}^{{{\tan }^{-1}}x}}+x{{e}^{{{\tan }^{-1}}x}}}{1+{{x}^{2}}} \right)dx $$
From equation (1), we get
$$ I=\int{1dt} $$
$$ I=t+C $$
On putting the value of $$t$$, we get
$$I=x{{e}^{{{\tan }^{-1}}x}}+C$$
Hence, this is the answer.
Consider the given integral.
$$I=\int{\dfrac{x}{\sqrt{{{x}^{2}}+2}}dx}$$
Let $$t={{x}^{2}}+2$$
$$ \dfrac{dt}{dx}=2x+0 $$
$$ \dfrac{dt}{2}=xdx $$
Therefore,
$$ I=\dfrac{1}{2}\int{\dfrac{dt}{\sqrt{t}}} $$
$$ I=\dfrac{1}{2}\left( 2\sqrt{t} \right)+C $$
$$ I=\sqrt{t}+C $$
$$I=\sqrt{{{x}^{2}}+2}+C$$
Consider the following integral.
$$ \int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx $$
Solution:
Let and differentiate w.r.t “x” we get.
$$ t={{2}^{x}} $$
$$ \dfrac{dt}{dx}={{2}^{x}}\ln 2 $$
$$ \int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right)\ $$
Put the value \[dx\,\,\] of in eq . (1) we get.
$$ =\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{\sqrt{1-{{4}^{x}}}}}dx\,\,\,\,\,\,\,\,......\left( 1 \right) $$
$$ =\int\limits_{{}}^{{}}{\dfrac{{{2}^{x}}}{{{2}^{x}}\ln 2\sqrt{1-{{\left( {{t}^{2}} \right)}^{x}}}}}dx $$
$$ =\int\limits_{{}}^{{}}{\dfrac{1}{\ln 2\sqrt{1-{{\left( {{t}^{2}} \right)}^{x}}}}}dx\, $$
$$ =\dfrac{1}{\ln 2}{{\sin }^{-1}}x+C $$
Hence, the value of $$[K=\dfrac{1}{\ln 2}]$$
Hence, this is the correct answer .
$$I=\int{\cos \left( \log x \right)dx}$$
Let $$t=\log x$$
$$ \dfrac{dt}{dx}=\dfrac{1}{x} $$
$$ xdt=dx $$
$$ I=\int{{{e}^{t}}\cos tdt} $$
$$ I=\cos t{{e}^{t}}-\int{\left( -\sin t \right){{e}^{t}}}dt $$
$$ I={{e}^{t}}\cos t+\int{\left( \sin t \right){{e}^{t}}}dt $$
$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-\int{\cos t{{e}^{t}}}dt $$
$$ I={{e}^{t}}\cos t+\sin t{{e}^{t}}-I $$
$$ 2I={{e}^{t}}\left( \cos t+\sin t \right)+C $$
$$ I=\dfrac{{{e}^{t}}}{2}\left( \cos t+\sin t \right)+C $$
On putting the value of $$'t'$$, we get
$$ I=\dfrac{{{e}^{\log x}}}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$
$$ I=\dfrac{x}{2}\left( \cos \left( \log x \right)+\sin \left( \log x \right) \right)+C $$
Consider the following integral .
$$ I=\int xsinxse{{c}^{3}}xdx=\int\limits_{{}}^{{}}{x\tan x{{\sec }^{2}}x} $$
let $$ t=\tan x $$ and differentiate both side w.r.t x, we get.
$$ \dfrac{dt}{{{\sec }^{2}}x}=dx $$
$$ =\int\limits_{{}}^{{}}{\dfrac{t{{\tan }^{-1}}t{{\sec }^{2}}x}{{{\sec }^{2}}x}}dt $$
$$ =\int\limits_{{}}^{{}}{t{{\tan }^{-1}}t}dt $$
$$ ={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\int\limits_{{}}^{{}}{\dfrac{1}{1+{{t}^{2}}}}}{{t}^{2}}dt $$
$$ ={{\tan }^{-1}}t\int\limits_{{}}^{{}}{tdt-\dfrac{1}{2}\left( \int\limits_{{}}^{{}}{\dfrac{{{t}^{2}}+1-1}{1+{{t}^{2}}}dt} \right)} $$
$$ =\dfrac{{{\tan }^{2}}x}{2}{{\tan }^{-1}}\left( \tan x \right)+\dfrac{{{\tan }^{2}}x}{2}{{\tan }^{-1}}\left( \tan x \right)-\dfrac{\tan x}{2}+C $$
$$ ={{\tan }^{2}}x{{\tan }^{-1}}\left( \tan x \right)-\dfrac{\tan x}{2}+C $$
Hence, this is the correct answer.
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