∫ limits1∞ ((x-1)4/x8+x10) d x is equal to

Question

∫ limits1∞ ((x-1)4/x8+x10) d x is equal to (A) π (B) (π/2) (C) 0 (D) None

Tardigrade Admin · Accepted Answer

Let I=∫ limits1∞ ((x-1)4/x8+x10) d x Putting x=(1/t), we get I =∫ limits10 (((1/t)-1)4/(1)t8+(1)t00(-(1/t2) d t) =∫ limits/0)1 (t4(1-t)4/(t2+1)) d t=(22/7)-π

Q. $1 \int \infty \frac{( x - 1 ) ^{4}}{x ^{8} + x ^{10}} d x$ is equal to

$π$

$\frac{π}{2}$

$0$

None

Let $I = 1 \int \infty \frac{( x - 1 ) ^{4}}{x ^{8} + x ^{10}} d x$
Putting $x = \frac{1}{t}$ , we get
$I = 1 \int 0 \frac{( \frac{1}{t} - 1 ) ^{4}}{\frac{1}{t ^{8}} + \frac{1}{t ^{00}}} (- \frac{1}{t ^{2}} d t)$
$= 0 \int 1 \frac{t ^{4} ( 1 - t ) ^{4}}{( t ^{2} + 1 )} d t = \frac{22}{7} - π$

Q. 1∫∞​x8+x10(x−1)4​dx is equal to

π

2π​

0

None

Let I=1∫∞​x8+x10(x−1)4​dx Putting x=t1​, we get I=1∫0​t81​+t001​(t1​−1)4​(−t21​dt) =0∫1​(t2+1)t4(1−t)4​dt=722​−π

Q. $1 \int \infty \frac{( x - 1 ) ^{4}}{x ^{8} + x ^{10}} d x$ is equal to

$π$

$\frac{π}{2}$

$0$

Let $I = 1 \int \infty \frac{( x - 1 ) ^{4}}{x ^{8} + x ^{10}} d x$
Putting $x = \frac{1}{t}$ , we get
$I = 1 \int 0 \frac{( \frac{1}{t} - 1 ) ^{4}}{\frac{1}{t ^{8}} + \frac{1}{t ^{00}}} (- \frac{1}{t ^{2}} d t)$
$= 0 \int 1 \frac{t ^{4} ( 1 - t ) ^{4}}{( t ^{2} + 1 )} d t = \frac{22}{7} - π$