-11-x3-x-3-x2-4-x-3 d x isA- 4-0-2ln sin- d -B- -4-0-2ln cos- d -C- -2-0-2ln sin 2 - d -D- -4-0-2ln sin-ln cos- d -

The correct option is B 4/π∫^π/2_0 ln (cosα) d α1 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 ...

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Byju's Answer

Standard XII

Mathematics

Chain Rule

-11∫x3+|x|+3/...

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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Similar questions

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1∫−1x3+|x|+3x2+4|x|+3dx is

The correct option is B −4π∫π20ln(cosα)dα1∫−1x3(|x|+1)(|x|+3)dx+1∫−1|x|+3(|x|+1)(|x|+3)dx =0+2∫101x+1dx=2 ln2 Now π/2∫0ln(sinα)dα=π/2∫0ln(cosα)dα=π/2∫0ln(sin2α)dα=−π2ln2

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