Mathematics

Integrate : $$\displaystyle\int _ { 1 } ^ { 2 } \frac { \ln x } { x ^ { 2 } } d x$$


SOLUTION
$$\Rightarrow \ \displaystyle \int_{1}^{2}\dfrac {lnx}{x^2}dx$$

let $$\ln x=t$$

$$\dfrac {1}{x}=\dfrac {dt}{dx}$$

$$\dfrac {1}{x}dx=dt$$

$$I=\displaystyle \int_{1}^{2}\dfrac {lnx}{x^2}dx=\displaystyle \int_{|n|=0}^{ln2}\dfrac {te^t}{e^{2t}}dt=\displaystyle \int_{0}^{ln2} t^{e^{-t}}dt$$

Applying By parts gives

$$I=(-te^{-t})_{0}^{ln2}+\displaystyle \int _{0}^{ln2}e^{-t}dt$$

$$I=-\ln2 \left (\dfrac {1}{2}\right)+(-e^{-t})_0^{ln2}$$

$$I=\dfrac { -\ln2 }{ 2 } +\left[ \dfrac { -1 }{ 2 } +1 \right] =\dfrac { 1 }{ 2 } -\dfrac { \ln2 }{ 2 } $$

$$\therefore \ \displaystyle \int_1^2 \dfrac {\ln2}{x^2}dx=\dfrac {1-\ln2}{2}$$
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Subjective Medium Published on 17th 09, 2020
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