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integrate ... x^3/sqrt(x^2-5)
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\[u=x^2-5=> du=2x dx \text{ and } u=x^2-5=>u+5=x^2\]
\[\text{ write } x^3 \text{ as } x^2 \cdot x \text{ first}\]
how do u do it? i think its other style...
\[\int\limits_{}^{}\frac{x^2 \cdot x}{\sqrt{x^2-5}} dx\]
yes i got that..
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\[\int\limits_{}^{}\frac{(u+5)}{\sqrt{u}} \frac{du}{2}\]
\[\frac{1}{2} \int\limits_{}^{}u^\frac{-1}{2}(u+5) du\]
why u+5/sqrt u??
\[x^2=u+5; x dx =\frac{du}{2}; x^2-5=u\]
|dw:1326765483867:dw|
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