The value of the integral 1∫0√xdx(1+x)(1+3x)(3+x) is
A
π4(1−√36)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
π8(1−√32)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
π4(1−√32)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
π8(1−√36)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Bπ8(1−√32) Put x=t2 1∫02t2(1+t2)(1+3t2)(3+t2)dt 2t2(1+t2)(1+3t2)(3+t2)=A1+t2+B1+3t2+C3+t2 ⇒A=12,B=−38,C=−38 ⇒121∫0dt1+t2−381∫0dt1+3t2−381∫0dt3+t2 =π8−38(1√3⋅π3)−38(1√3⋅π6) =π8−√316π=π8(1−√32)