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help please integral of sqrt (x^2+6x) dx
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\[\int\limits \sqrt(x^2+6x) dx\]
i dunno, there seems to be alot of variables to account for ....
its really hard
\[\int\limits_{}^{}\sqrt{x^2+6x+9-9} dx=\int\limits_{}^{}\sqrt{(x+3)^2-3^2} dx\] you will need a trig substitution
(x^(2)+6x)*dx Multiply dx by each term inside the parentheses. dx^(3)+6dx^(2)
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