1-1-x3 - -x- - 3-x2 - 4-x- - 3dx is

1 1 nbsp;∫x^3/(|x| + 1)(|x| + 3)dx + 1 1 nbsp;∫|x| + 3/(|x| + 1)(|x| + 3)dx = 0 + 2 ∫^1_01/x+1dx = 2 ln 2 Now π/20 nbsp;∫ nbsp;ln (sinα)d α = π/20 nbs ...

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Byju's Answer

Standard XI

Mathematics

Chain Rule

1-1∫x3 + |x| ...

Question

$1 \int - 1 \frac{x^{3} + | x | + 3}{x^{2} + 4 | x | + 3} d x$ is

\frac{4}{π} \int_{0}^{\frac{π}{2}} ln (sin α) d α

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- \frac{4}{π} \int_{0}^{\frac{π}{2}} ln (cos α) d α

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- \frac{2}{π} \int_{0}^{\frac{π}{2}} ln (sin 2 α) d α

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- \frac{4}{π} \int_{0}^{\frac{π}{2}} (ln (sin α) + ln (cos α)) d α

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Solution

The correct option is B $- \frac{4}{π} \int_{0}^{\frac{π}{2}} ln (cos α) d α$
$1 \int - 1 \frac{x^{3}}{(| x | + 1) (| x | + 3)} d x + 1 \int - 1 \frac{| x | + 3}{(| x | + 1) (| x | + 3)} d x$
$= 0 + 2 \int_{0}^{1} \frac{1}{x + 1} d x = 2 ln 2$
Now $π / 2 \int 0 ln (sin α) d α = π / 2 \int 0 ln (cos α) d α = π / 2 \int 0 ln ({sin}^{2} α) d α = - \frac{π}{2} ln 2$

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1∫−1x3+|x|+3x2+4|x|+3dx is

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$1 \int - 1 \frac{x^{3} + | x | + 3}{x^{2} + 4 | x | + 3} d x$ is