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Evaluate the indefinite integral x^3sqrt(12+x^4)dx
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let u=12+x^4 du=4x^3 dx
you got the answer i was confused on what to integrate first the sqrt(u) or just (u) itself because i kept on getting the wrong answer
if du=4x^3 dx then du/4=x^3 dx right?
\[\int\limits_{}^{}x^3 \sqrt{12+x^4} dx =\int\limits_{}^{} \sqrt{12+x^4} x^3 dx\] \[=\int\limits_{}^{}\sqrt{u} \frac{du}{4}\]
\[=\frac{1}{4}\int\limits_{}^{}\sqrt{u} du=\frac{1}{4}\int\limits_{}^{}u^\frac{1}{2} du\]
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yeah i got eh right answer in the end but this step by step gives a better understanding too
\[=\frac{1}{4} \cdot \frac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C\] replace u with 12+x^4
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