Question

Please solve this ASAP
and please no fake answers ​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Concept}}}$

Here the concept of integration has been used. Firstly we can assume different terms given here as variables. Then we can apply them in the given expression and finally integrate them to get the answer.

Let's do it !!

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★ Solution :-

Given expression ::

$\\\;\displaystyle{\bf{\mapsto\;\;\blue{\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx}}}$

Now,

• log x = t

$\\\;\sf{\leadsto\;\;Then\;here,\;x\;=\;\pink{e^{t}}}$

Combining both these equations and differentiating, we get

$\\\;\bf{\rightarrow\;\;\orange{dx\;=\;e^{t}\:dt}}$

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~ For Final Answer :-

We are given that,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx}}$

This can be written as,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\int\;\:\bigg[\dfrac{1}{t}\;-\;\dfrac{1}{t^{2}}\bigg]\;e^{t}dt}}}$

Now we can take f(x) = 1/t and f'(x) = - 1/t²

By applying this, we get

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\int\;\:\bigg[f(x)\;-\;f'(x)}\bigg]\;e^{t}dt}}$

On differentiating this, we get

$\\\;\rm{\mapsto\;\;\red{\dfrac{1}{dt}\bigg(\dfrac{1}{t}\bigg)\;=\;-\:\dfrac{1}{t^{2}}}}$

This will give us,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{e^{t}\;\times\:\dfrac{1}{t}\;+\;c}}}$

This shows the integral property.

Now we can substitute the value of t as log x, we get

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{e^{log\:x}\;\times\:\dfrac{1}{log\:x}\;+\;c}}}$

Then this will give,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{x\;\times\:\dfrac{1}{log\:x}\;+\;c}}}$

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\purple{\dfrac{x}{log\:x}\;+\;c}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\;integral\;=\;\bf{\purple{\dfrac{x}{log\:x}\;+\;c}}}}}$

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★ More to know :-

• Power rule of integration (n ≠ -1) :-

$\\\;\displaystyle{\tt{\leadsto\;\;\int\;x^{n}dx\;=\;\dfrac{x^{n\:+\:1}}{n\;+\;1}\;+\;c}}$

• Sum Rule :-

$\\\;\displaystyle{\tt{\leadsto\;\;\int\;(f\;+\;g)dx\;=\;\int\:f\;dx\;+\;\int\:g\;dx}}$

• Difference Rule :-

$\\\;\displaystyle{\tt{\leadsto\;\;\int\;(f\;-\;g)dx\;=\;\int\:f\;dx\;-\;\int\:g\;dx}}$

• Multiplication by constant :-

$\\\;\displaystyle{\tt{\leadsto\;\;cf(x)\:dx\;=\;c\int\:f(x)\:dx}}$

Answer 2

$\Huge{\mathfrak{\purple{\underline{Answer}}}}$

Given expression ::

$\\\;\displaystyle{\bf{\mapsto\;\;\blue{\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx}}}$

Now,

log x = t

$\\\;\sf{\leadsto\;\;Then\;here,\;x\;=\;\pink{e^{t}}}$

Combining both these equations and differentiating, we get

$\\\;\bf{\rightarrow\;\;\orange{dx\;=\;e^{t}\:dt}}$

~ For Final Answer :-

We are given that,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx}}$

This can be written as,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\int\;\:\bigg[\dfrac{1}{t}\;-\;\dfrac{1}{t^{2}}\bigg]\;e^{t}dt}}}$

Now we can take f(x) = 1/t and f'(x) = - 1/t²

By applying this, we get

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\int\;\:\bigg[f(x)\;-\;f'(x)}\bigg]\;e^{t}dt}}$

On differentiating this, we get

$\\\;\rm{\mapsto\;\;\red{\dfrac{1}{dt}\bigg(\dfrac{1}{t}\bigg)\;=\;-\:\dfrac{1}{t^{2}}}}$

This will give us,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{e^{t}\;\times\:\dfrac{1}{t}\;+\;c}}}$

This shows the integral property.

Now we can substitute the value of t as log x, we get

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{e^{log\:x}\;\times\:\dfrac{1}{log\:x}\;+\;c}}}$

Then this will give,

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{x\;\times\:\dfrac{1}{log\:x}\;+\;c}}}$

$\\\;\displaystyle{\sf{\Longrightarrow\;\;\int\;\:\bigg[\dfrac{1}{log\:x}\;-\;\dfrac{1}{(log\:x)^{2}}\bigg]\;dx\;=\;\bf{\purple{\dfrac{x}{log\:x}\;+\;c}}}}$

This is the required answer.

$\\\;\underline{\boxed{\tt{Required\;\;integral\;=\;\bf{\purple{\dfrac{x}{log\:x}\;+\;c}}}}}$