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

Integrals of the form R(x,ax2+bx+c) dx are calculated with the aid of one of the three Euler substitutions.
I. ax2+bx+c=t±xa if a>0;
II. ax2+bx+c=tx±c if c>0;
III. ax2+bx+c=(xa)t±x if ax2+bx+c=a(xα)(xβ) i.e., if α is a real root of ax2+bx+c=0.

x dx(7x10x2) can be evaluated by substituting for x as

A
x=5+2t2t2+1
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
x=5t2t2+2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
x=2t253t21
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 x=5+2t2t2+1
In this case, a<0 and c<0.
Therefore, neither the first nor the second Euler substitution is applicable. But the quadratic 7x10x2 has real roots α=2, β=5.
Therefore, we use the third Euler substitution:
7x10x2=(x2)(5x)=(x2)t
5x=(x2)t2
x=5+2t2t2+1

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Integration by Partial Fractions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon