Q1 \int_{a}^{b}\frac{e^{x/a}-e^{b/x}}{x}dx
Q2 \int_{0}^{1}\frac{lnx.ln(1-x)}{(1+x)^2}dx
241st one is done....now everyone plzz try 2nd one
and uttara i couldnt understand ur soln at all
66Q1)
I=\int_a^b \dfrac{e^{x/a}-e^{b/x}}{x}\ \mathrm dx
Substitute x = ay giving dx =a dy. Also when x=a, y =1, and when x =b, y = b/a
So
I=\int_1^{b/a} \dfrac{e^y-e^{b/ay}}{y}\ \mathrm dy
Next substitute y = bat which gives dy = - ba t2 dt. Also when y =1, t =b/a and when y =b/a, t=1. So
I=\int_1^{b/a} \dfrac{e^{b/at}-e^{t}}{t}\ \mathrm dt = -\int_1^{b/a} \dfrac{e^{t}-e^{b/at}}{t}\ \mathrm dt =-I
Hence, I=0
24Thanks sir..
but i used e^{x/a}=\sum_{i=0}^{\infty}{x^i/a^i.i!} \ and \ e^{b/x}=\sum_{i=0}^{\infty}{b^i/x^i.i!}
though that way it was a bit long [3]
66Q2)
I=\int_0^1\dfrac{\ln x\ln(1-x)}{(1+x)^2}\ \mathrm dx
Integration by parts gives us
I=\left.-\dfrac{\ln x\ln(1-x)}{1+x}\right|_0^1+\int_0^1\left(\dfrac{\ln(1-x)}{x(1+x)}-\dfrac{\ln x}{1-x^2}\right)\mathrm dx
The first part evaluates to zero (Can you show that?), while the remaining integral becomes
I=\int_0^1\left(\dfrac{\ln(1-x)}{x}-\dfrac{\ln(1-x)}{1+x}\right)\mathrm dx-\int_0^1\dfrac{\ln x}{1-x^2}\right)\mathrm dx
I=\int_0^1\dfrac{\ln(1-x)}{x}\ \mathrm dx-\int_0^1\dfrac{\ln(1-x)}{1+x} \ \mathrm dx-\int_0^1\dfrac{\ln x}{1-x^2}\right)\mathrm dx
I=I_1-I_2-I_3
Next, we have
I=I_1-I_3=\int_0^1\left(\dfrac{\ln x}{1-x}-\dfrac{\ln x}{1-x^2}\right)\ \mathrm dx=\int_0^1\dfrac{x\ln x}{1-x^2}\ \mathrm dx
Substitute x = √t so that dx = 12√t dt. The above integral reduces to
\int_0^1\dfrac{x\ln x}{1-x^2}\ \mathrm dx=\dfrac{1}{4}\int_0^1\dfrac{\ln t}{1-t}\ \mathrm dt
Next use the following results:
1) \int_0^1\dfrac{\ln x}{1-x}\ \mathrm dx=-\dfrac{\pi^2}{6} and
2) \int_0^1\dfrac{\ln (1-x)}{1+x}\ \mathrm dx=\dfrac{\ln^22}{2}-\dfrac{\pi^2}{12}
To get
\boxed{I=\dfrac{\pi^2}{24}-\dfrac{\ln^22}{2}}
P.S. I prove the results 1) and 2) in the following.
4I used integration by parts but din't know how to proceed
thanks to Anant Sir