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Evaluate the integral 7x/ x squared +2 [1,5]
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\[\int\limits_{1}^{5} 7x/ x^2+2 dx\]
make the substitution \(x^2+2=u\)
then \[2xdx=du\implies x\,dx={du\over2}\] when \(x=1,u=3\) when \(x=5,u=27\)
Get the integral of the function and solve for x at 5 and 2
i mean 5 and 1
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solve the integral in u, replace u with \(x^2+2\) and apply the limits.
\[ \int_3^{27}{7\over u}{du\over2}={7\over2}[\ln|u|]_3^{27}={7\over2}[\ln(27-\ln(3)]\\ \quad={7\over2}[3\ln(3)-\ln(3)]=\mathbf{7\ln(3)} \]
I find this way easier
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