wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

Integrate the following functions.
176xx2dx.

Open in App
Solution

Let I=176xx2dx=17(x2+6x)dx
=17[x2+2×3x+(3)2(3)2]dx=17[x2+6x+329]dx=17[(x+3)29]dx=142(x+3)2dx
Let x+3=tdx=dt
I=142t2dt=sin1(t4)+C[dxa2x2=1asin1(xa)]=sin1(x+34)+C(t=x+3)


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