Question #132614
∫ t(2t-3√t)dt
1
Expert's answer
2020-09-13T18:22:02-0400

Let us rewritet(2t3t)dt\int t (2 t - 3 \sqrt{t}) dt as (2t23t3/2)dt\int (2 t^2 - 3 t^{3/2}) dt.

Using table integral xndx=xn+1n+1+C\int x^n dx = \frac{x^{n+1}}{n+1} + C, obtain:

(2t23t3/2)dt=2t33325t5/2+C=23t365t5/2+C\int (2 t^2 - 3 t^{3/2}) dt = 2 \frac{t^3}{3} - 3 \cdot \frac{2}{5} t^{5/2} + C = \frac{2}{3}t^{3} - \frac{6}{5} t^{5/2} + C.


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