-1-ln 2 x2-1-ln 2 x2 x-1-ln 2 x-2 x2 d x is equal to where c is constant of integrationA- ln -1-x ln x-cB- ln -1-x ln x-cC- 1-2ln -1-x ln x-cD- 1-2ln -1-2 x ln 2 x - c

The correct option is D 1/2 ln |1 + 2x ln 2x| + c∫(1+ln 2x)^21 + (2x+1)ln 2x + 2x (ln 2x)^2dx = ∫(1+ln 2x)^2(1+2x ln 2x) (1+ln 2x) dx 1 + 2x ln 2x = t 2(1+l ...

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Byju's Answer

Standard XII

Mathematics

Second Fundamental Theorem of Calculus

∫1+ln 2 x2/1+...

Question

$\int \frac{(1 + ln 2 x)^{2}}{1 + ln (2 x)^{2 x + 1} + (ln (2 x)^{\sqrt{2 x}})^{2}} d x$ is equal to (where $c$ is constant of integration)

ln | 1 + x ln x | + c

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\frac{1}{2} ln | 1 + x ln x | + c

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\frac{1}{2} ln | 1 + 2 x ln 2 x | + c

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ln | 1 - x ln x | + c

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Solution

The correct option is C $\frac{1}{2} ln | 1 + 2 x ln 2 x | + c$
$\int \frac{(1 + ln 2 x)^{2}}{1 + (2 x + 1) ln 2 x + 2 x (ln 2 x)^{2}} d x = \int \frac{(1 + ln 2 x)^{2}}{(1 + 2 x ln 2 x) (1 + ln 2 x)} d x 1 + 2 x ln 2 x = t 2 (1 + ln 2 x) d x = d t = \frac{1}{2} \int \frac{d t}{t} = \frac{1}{2} ln t + c = \frac{1}{2} ln | 1 + 2 x ln 2 x | + c$

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∫(1+ln2x)21+ln(2x)2x+1+(ln(2x)√2x)2 dx is equal to (where c is constant of integration)

The correct option is C 12 ln|1+2x ln 2x|+c∫(1+ln2x)21+(2x+1)ln2x+2x(ln2x)2dx=∫(1+ln2x)2(1+2xln2x)(1+ln2x)dx1+2xln2x=t2(1+ln2x)dx=dt=12∫dtt=12lnt+c=12ln|1+2x ln2x|+c

$\int \frac{(1 + ln 2 x)^{2}}{1 + ln (2 x)^{2 x + 1} + (ln (2 x)^{\sqrt{2 x}})^{2}} d x$ is equal to (where $c$ is constant of integration)