Given that we may write x/(x+4)^2 as (x+4)-4/(x+4)^2, find the integral (limits 1 and 0) of x/(x+4)^2 dx. Thank you
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Originally Posted by slaypullingcat Given that we may write x/(x+4)^2 as (x+4)-4/(x+4)^2, find the integral (limits 1 and 0) of x/(x+4)^2 dx. Thank you $\displaystyle \int_0^1\frac x{(x+4)^2}\,dx=\int_0^1\frac{x+4-4}{(x+4)^2}\,dx$ $\displaystyle =\int_0^1\frac{dx}{x+4}-4\int_0^1\frac{dx}{(x+4)^2}$ Use the log rule for the first integral, and the power rule for the second.
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