1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# The value of the integral ∫x3+x+1x2−1dx is (where C is an arbitrary constant)

A
x22+ln|x+1|+32ln|x1|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
x22+12ln|x+1|+32ln|x1|+C
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
x22ln|x+1|+23ln|x1|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
x22+ln|x1|+32ln|x+1|+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is B x22+12ln|x+1|+32ln|x−1|+CIt can be seen that the given integrand is not a proper rational expression, therefore, on dividing x3+x+1x2−1=x+2x+1x2−1 Let 2x+1x2−1=A(x+1)+B(x−1)⇒2x+1=A(x−1)+B(x+1)⋯(1) Equating the coefficients of x and constant, we obtain A+B=2−A+B=1 On solving, we obtain A=12 and B=32 ∴x3+x+1x2−1=x+12(x+1)+32(x−1)⇒∫x3+x+1x2+1dx=∫x dx+12∫1(x+1)dx+32∫1(x−1)dx=x22+12ln|x+1|+32ln|x−1|+C

Suggest Corrections
1
Join BYJU'S Learning Program
Join BYJU'S Learning Program