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

Evaluate the following integrals:
x3-3xx4+2x2-4dx

Open in App
Solution

I=x3-3xx4+2x2-4dx

=x(x2 -3)x4+2x2-4dx

Let x2=t, or, 2xdx=dt

I=12(t-3)t2+2t-4dt=142t-6t2+2t-4dt=142t+2-8t2+2t-4dt=142t+2t2+2t-4-8t2+2t-4dt=142t+2t2+2t-4dt-8t2+2t-4dt

I=14I1+I2 ...i

Now,

I1 = 2t + 2t2 + 2t -4 dt

Let t2+2t-4=u

or, 2t+2dt=duI1=1u du=lnu+c1I1=lnt2+2t-4+c1 I1=lnx4+2x2-4+ c1

Now,

I2=-8(t+1)2-5dtI2=8(5)2-(t+1)2dt I2=825ln5+x2+15-x2-1+c2

So,from i, we getI=14lnx4+2x2-4+45ln 5+x2+15-x2-1+C I=14lnx4+2x2-4+15ln 5+x2+15-x2-1+C

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Property 4
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon