Ad Infinitum!!

This 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

let (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

\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??

no diff

So everyone writes ∂Q/∂x just to sound more funda-like?? [3]