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\[\int\limits_{0}^{2}\left(\begin{matrix}3x \\ x ^{2}+4\end{matrix}\right)\]
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\[\int\limits\frac{3x}{x^{2}-4} dx\] Not a matrices?
\[\int\limits\frac{3x}{x^{2}+4} dx\]
yup. :) but it is a definite integration. :)
let \(u=x^2+4\) => \[dx= \frac{du}{2x} \]
answer? :)
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answer @Mimi_x3 ? :)
\[\int\limits\frac{3x}{u} * \frac{du}{2x} => \int\limits\frac{3\cancel{x}}{u} * \frac{du}{2\cancel{x}} => \frac{3}{2} \int\limits\frac{1}{u} du\]
you may take 3 out then take 1/2 common such that 1/2 is out and inside there is 2x now x^2 + 4 =t so 2xdx = dt => 3/2 (dt/t) try now..
just what @Mimi_x3 says :-)
Integrate that then don't forget to sub back in the \(u\). (:
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ok thanks @Mimi_x3 :)
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