plz solve for me in detail this question:Q.integrate 2^x.e^x wrt x

Hi Nishad,

Let I = ∫2^xe^xdx.

Integrate by parts, we get

I = 2^xe^x - ∫2^xe^xlog2dx + C

So I = 2^xe^x - log2*I + C

So (1+log2)I = 2^xe^x + C

So I = 2^xe^x/(1+log2) + C'

Or I = 2^xe^x/log(2e) + C'

That is the integral of the above function.

Best Regards,

Ashwin (IIT Madras).

Dear nishad bagwe,

The Integration of ∫2xe^x dx is 2[e^x(x-1) + c]

Let I = 2∫x.e^x dx

      Using the integration by parts method,

Take u = x               v =e^x dx

        du = dx           dv=e^x

Now sustitute them by the formula,

∫udv = uv - ∫vdu

Now you will get this answer .

Hope this helped you immensely..

All the Very Best & Good Luck to you ..

Regards,

AskIITians Expert

Godfrey Classic Prince

IIT-M

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