Question

Calcule a integral

k) $\int\ {}(3u^5-2u^3) \, du$

Answer 1

Oi Lucas

√ (3u⁵ - 2u³) du = √ 3u⁵ du - √ 2u³ du 

√ 3u⁵ du - √ 2u³ du = 3*u^(⁵+¹)/(5 + 1) - 2*u^(³+¹)/(3 + 1) 

√ 3u⁵ du - √ 2u³ du = u⁶/2 - u⁴/2 + C 

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Answer 2

$\sf \int \left(3u^5-2u^3\right)du\\\\\\=\int \:3u^5du-\int \:2u^3du\\\\\\=\dfrac{u^6}{2}-\dfrac{u^4}{2}\\\\\\\to \boxed{\sf =\frac{u^6}{2}-\frac{u^4}{2}+C}$