IIT JEE , AIEEE Forum - Mathematics , Integral Calculus

ankurgupta91

4) put ln(ln(lnx)) = t
dx/ (xln(lnx) lnx) = dt
then
=integral tdt
= t^2/2

nwu wl get the answer............

ankurgupta91

1) integral [2^(x+1) -3^(x-1)] / (2^x.3^x)

integral 2/ (3^x) - 1/ 3.(2^x)

= -2ln(3) / 3^x + ln(2) / 3.(2^x)

thats the answer..........

sachinguptaiit

I= $\int$ a^xb^xc^x.dx

I=a^xb^x $\int$ c^xdx - $\int$ a^xb^x(loga+logb).(c^x/logc)dx

I=(a^xb^xc^x/log(c))- $\int$ a^xb^xc^x(log(ab)/log(c))dx

I=(a^xb^xc^x/log(c))-I.(log_cab)

I= (a^xb^xc^x)/(log(c).(1+log_cab))=(a^xb^xc^x/log(abc))

Note:- For those who cannot understand this solution,I will type out solution in LaTeX and provide them

sachinguptaiit

I = $\int$ (2.2^x-(1/3).3^x)/(2^x.3^x)dx

I = 2 $\int$ 3^-xdx- (1/3) $\int$ 2^-xdx

I = -2(1/(3^x.ln3))+(1/(3.2^x.ln2))+C

kaymant

Sachin,

Was there any need to use integration by parts?

$\int a^x b^x c^x\ \mathrm{d}x =\int (abc)^x\ \mathrm{d}x = \dfrac{(abc)^x}{\ln(abc)} +C$

reddevil_2009

gr8888888 sir.................one line solution.......................

gx

4) put lnx=t

1/xdx=dt

so ln(lnt) dt/t*lnt

put lnt=z

so it comes to lnz dz/z

hope u can integrate it now