How do I evaluate #intz/e^(3z) dz#?

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1 Answer
Gió
Jan 29, 2015

I would first write your integral in a more friendly form:
#intz/(e^(3z))dz=intz*e^(-3z)dz#

I would then use integration by parts:
Where you have:
#intf(z)*g(z)dz=F(z)*g(z)-intF(z)*g'(z)dz#

Where:
#F(z)=intf(z)dz#
#g'(z)# is the derivative of #g(z)#

In your case you can choose:
#f(z)=e^(-3z)#
#g(z)=z#
And:
#intz*e^(-3z)dz=z*e^(-3z)/(-3)-int1*e^(-3z)/(-3)dz=#
#=z*e^(-3z)/(-3)-e^(-3z)/9+c#

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