-log x2 d x-A- xlog x2-2 x log x-2 x-cB- xlog x2-2 x log x-2 x-cC- xlog x2-2 x log x-x-cD- xlog x2-2 x log x-x-c

The correct option is B x(log x)^2 2x log x+2x+c∫ (log x)^2 dx Put log x=t ⇒ e^t=x⇒ dx=e^t dt, then it reduces to ∫ t^2.e^t dt=t^2 e^t 2 te^t+2e^t + c =x(log ...

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Byju's Answer

Standard XII

Mathematics

Bijective Function

∫log x2 d x=A...

Question

$\int (l o g x)^{2} d x =$

x (l o g x)^{2} - 2 x l o g x - 2 x + c

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x (l o g x)^{2} - 2 x l o g x - x + c

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x (l o g x)^{2} - 2 x l o g x + 2 x + c

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x (l o g x)^{2} - 2 x l o g x + x + c

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Solution

The correct option is C $x (l o g x)^{2} - 2 x l o g x + 2 x + c$
$\int (l o g x)^{2} d x$
Put $l o g x = t \Rightarrow e^{t} = x \Rightarrow d x = e^{t} d t$ , then it reduces to $\int t^{2} . e^{t} d t = t^{2} e^{t} - 2 t e^{t} + 2 e^{t} + c = x (l o g x)^{2} - 2 x l o g x + 2 x + c$

Suggest Corrections

∫(log x)2 dx=

The correct option is C x(log x)2−2x log x+2x+c∫(log x)2 dx Put log x=t⇒et=x⇒dx=et dt, then it reduces to ∫t2.et dt=t2et−2 tet+2et+c=x(log x)2−2x log x+2x+c

$\int (l o g x)^{2} d x =$

The correct option is C $x (l o g x)^{2} - 2 x l o g x + 2 x + c$
$\int (l o g x)^{2} d x$
Put $l o g x = t \Rightarrow e^{t} = x \Rightarrow d x = e^{t} d t$ , then it reduces to $\int t^{2} . e^{t} d t = t^{2} e^{t} - 2 t e^{t} + 2 e^{t} + c = x (l o g x)^{2} - 2 x l o g x + 2 x + c$