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Find the indefinite integral of the following function: (2/ cubed root of t) + 3 The answer is: 3t^(2/3) + 3t + C but I need the steps because I clearly can't seem to figure it out!
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\[\huge \int{ \left(\frac{2}{t^{\frac{1}{3}}}+3 \right) dt } \rightarrow2\int{{t^{-\frac{1}{3}}}dt}+3\int{dt}\] \[\huge2\int{{t^{-\frac{1}{3}}}dt}\rightarrow2 \left[\frac{t^{-\frac{1}{3}+1}}{- \frac{1}{3}+1} \right] \rightarrow\]\[\huge 2 \left(\frac{t^{\frac{2}{3}}} {\frac{2}{3}} \right)+c \rightarrow 3t^{\frac{2}{3}}+c1 \] \[\huge 3\int{dt} \rightarrow 3t+c2\] C1+C2 = C All together \[\huge 2\int{{t^{-\frac{1}{3}}}dt}+3\int{dt} \rightarrow 3t^{\frac{2}{3}}+3t+C \]
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