CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
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
Or2t.tdt
= 2t2dt
= 2t33 +C
Let’s put t =x2+x+1
So, the final answer would be 2(x2+x+1)323 +C

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Special Integrals - 3
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon