∫x^3 (√(x^2 +4))dx


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Question Number 136304 by aurpeyz last updated on 20/Mar/21

$\int x^{3} \sqrt{x^{2} + 4} d x$
	Answered by liberty last updated on 20/Mar/21
	$∫ x^2 (x(√(x^2 +4)) ) dx { ((u=x^2 →du=2x dx)),((v=∫ x(√(x^2 +4)) dx = (1/3)(x^2 +4)^(3/2) )) :} I=(1/3)x^2 (x^2 +4)^(3/2) −(1/3)∫ (x^2 +4)^(3/2) d(x^2 +4) I=((x^2 (x^2 +4)^(3/2) )/3) − ((2(x^2 +4)^(5/2) )/(15)) + c$
	$\int x^{2} (x \sqrt{x^{2} + 4}) d x$ ${\begin{cases} u = x^{2} \to d u = 2 x d x \\ v = \int x \sqrt{x^{2} + 4} d x = \frac{1}{3} {(x^{2} + 4)}^{3 / 2} \end{cases}$ $I = \frac{1}{3} x^{2} {(x^{2} + 4)}^{3 / 2} - \frac{1}{3} \int {(x^{2} + 4)}^{3 / 2} d (x^{2} + 4)$ $I = \frac{x^{2} {(x^{2} + 4)}^{3 / 2}}{3} - \frac{2 {(x^{2} + 4)}^{5 / 2}}{15} + c$
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