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It is given that
$$
\frac{d}{d t}(t \log t-t)=\log t \text { then } \exp \left(\int_0^1 2 x \log \left(1+x^2\right) d x\right)=
$$
Options:
$$
\frac{d}{d t}(t \log t-t)=\log t \text { then } \exp \left(\int_0^1 2 x \log \left(1+x^2\right) d x\right)=
$$
Solution:
1750 Upvotes
Verified Answer
The correct answer is:
$\frac{4}{e}$
Given $\frac{d}{d t}(t \log t-t)=\log t$
Take, $\exp \left(\int_0^1 2 x \log \left(1+x^2\right) d x\right)$
Let $1+x^2=t$
$2 x d x=d t$
when $x=0, t=1$
$x=1, t=2$
Then, $\exp \left(\int_1^2 \log t-d t\right)=\left.\exp (t \log t-t)\right|_1 ^2$
$=\exp (2 \log 2-2-\log 1+1)$
$=\exp (\log 4-1)=e^{\ln 4} e^{-1}=\frac{4}{e}$
So, correct option is (c)
Take, $\exp \left(\int_0^1 2 x \log \left(1+x^2\right) d x\right)$
Let $1+x^2=t$
$2 x d x=d t$
when $x=0, t=1$
$x=1, t=2$
Then, $\exp \left(\int_1^2 \log t-d t\right)=\left.\exp (t \log t-t)\right|_1 ^2$
$=\exp (2 \log 2-2-\log 1+1)$
$=\exp (\log 4-1)=e^{\ln 4} e^{-1}=\frac{4}{e}$
So, correct option is (c)
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