If ∫x5e−4x3dx=148e−4x3f(x)+C, where C is a constant of integration, then f(x) is equal to :
A
−2x3−1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
−4x3−1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
4x3+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
−2x3+1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is B−4x3−1 Given, ∫x5e−4x3dx=148e−4x3f(x)+C
Put x3=t∴3x2dx=dt ⇒∫x3e−4x3x2dx=13∫te−4tdt =13[te−4t−4−∫e−4t−4dt] =−e−4t48[4t+1]+C =−e−4x348[4x3+1]+C ∴f(x)=−1−4x3