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

# ∫(2x+1)√x2+x+1 dx will be equal to -

A
2(x2+x+1)3/23+C

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2(x2+x+1)3/27+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
2(x2+x+1)3/29+C
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
None of these
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

## The correct option is A 2(x2+x+1)3/23+C The given question is ∫(px+q)√ax2+bx+c form.To approach these kind of questions we express px + q in terms of the derivative of the quadratic expression inside the under root. So, px + q = αd dx(ax2+bx+c)+β Where αandβ are the constants. Here, we can see the given linear expression is already the derivative of the quadratic expression given So, here α=1,β=0. ∫(2x+1)√x2+x+1 dx Let’s substitute x2+x+1=t2 So,(2x+1)dx = 2t dt =∫2t.√t2dt Or∫2t.tdt = ∫2t2dt = 2t33 +C Let’s put t =√x2+x+1 So, the final answer would be 2(x2+x+1)323 +C

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Special Integrals - 3
MATHEMATICS
Watch in App
Explore more
Join BYJU'S Learning Program