I have been sitting on this problem for an hr but it will be simple for u guys........ My brain hasn't been working too well after spending 5 days reading biology!!!
\int (3x-2)\sqrt{x^{2}+x+1}
21This is a typical question.
Ani : I = L√Q
wer L = linear; Q = quad
Express linear as t(∂Q/∂x).................... wer t = const. to adjust "3" of 3x
now ur I bcums
I = t * ∂Q/∂x * √Q + t' √Q .............. wer t' is the remain over of of ur original const. in L i.e. -2
NOW U CAN DO IT
1let (3x-2) = A.(d/dx(x2+x+1)+ B
then 3x-2=2Ax+A+B
then 2Ax=3x
A=3/2
A+B=-2
B=-7/2
therefore
(3x-2)=3/2(2x+10)-7/2
∫(3x-2)√x2+x+1 dx = ∫3/2(2x+1) -7/2∫√X2+X+1 DX
=3/2∫(2X+1)√X2+X+1 DX - 7/2 ∫√X2+X+1 DX
I THINK U CAN SOLVE FURTHER BY TAKING X2+X+1=T
11\int \left[ \frac{3}{2}(2x+1)-\frac{7}{2} \right]\sqrt{x^{2}+x+1}\: dx
So we have to express it like this and then proceed???
PS:
BTW, in what way is ∂Q/∂x different from dQ/dx??
11Sorry amit....didn't see ur post...
That question was to tapan
11So everyone writes ∂Q/∂x just to sound more funda-like?? [3]