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Example of Integration by Parts - Integration techniques
Illustration1: Evaluate the following integral.
∫ xe6x dx
Solution :
Thus, on some level, the difficulty here is the x that is in front of the exponential. If that was not there we could do the integral. Notice also, that in doing integration by parts anything that we wish for u will be differentiated. Thus, it seems that choosing u = x will be a good choice as upon differentiating the x will drop out.
Here that we've selected u we know that dv will be everything else which remains. Thus, here are the choices for u and dv also du and v.
u = x dv = e6x dx
du = dx v = e6x dx = 1/6e6x
Then the integral is as follow:
∫ xe6x dx = x/6 e6x - ∫ 1/6 e6x dx
= x/6 e6x - 1/36 e6x + c
Just once we have completed the last integral in the problem we will add in the constant of integration to obtain our final answer.
2 5 - 6 7 4 6 8 -10 8- 6 1 4
we know that log1 to any base =0 take antilog threfore a 0 =1
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