Question

Preciso de ajuda com esta integral, alguém sabe como resolvê-la??
∫(4,1) x^(3/2) ln(x)

Answer 1

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$\displaystyle\sf\int x^{\frac{3}{2}}\ell n(x)~dx\\\underline{\rm fac_{\!\!,}a}\\\sf u=\ell n(x)\implies du=\dfrac{1}{x}dx\\\sf dv=x^{\frac{3}{2}}~dx\implies v=\dfrac{2}{5}x^{\frac{5}{2}}\\\displaystyle\sf\int x^{\frac{3}{2}}\ell n(x)~dx=\ell n(x)\cdot\dfrac{2}{5}x^{\frac{5}{2}}-\dfrac{2}{5}\int x^{\frac{5}{2}}\cdot\dfrac{1}{x}~dx\\\displaystyle\sf\int x^{\frac{3}{2}}\ell n(x)~dx=\ell n(x)\cdot\dfrac{2}{5}x^{\frac{5}{2}}-\dfrac{2}{5}\int x^{\frac{3}{2}}~dx\\\displaystyle\sf\int x^{\frac{3}{2}}\ell n(x)~dx=\dfrac{2}{5}\ell n(x)\cdot x^{\frac{5}{2}}-\dfrac{2}{5}\cdot\dfrac{2}{5}\cdot x^{\frac{5}{2}}+c\\\displaystyle\sf\int x^{\frac{3}{2}}\ell n(x)=\dfrac{2}{5}\ell n(x)\cdot x^{\frac{5}{2}} -\dfrac{4}{25}x^{\frac{5}{2}}+c$

$\displaystyle\sf\int_{1}^{4}x^{\frac{3}{2}}\ell n(x)~dx=\bigg[\dfrac{2}{5}\ell n(x)\cdot x^{\frac{5}{2}}-\dfrac{4}{25}x^{\frac{5}{2}}\bigg]_{1}^{4}\\\sf=\dfrac{2}{5}\ell n(4)\cdot 4^{\frac{5}{2}}-\dfrac{4}{25}\cdot4^{\frac{5}{2}}-\bigg[\dfrac{2}{5}\ell n(1)\cdot1^{\frac{5}{2}}-\dfrac{4}{25}\cdot1^{\frac{5}{2}}\bigg]\\\sf =\dfrac{2}{5}\ell n(4)\cdot32-\dfrac{4}{25}\cdot32-\bigg[0-1\bigg]=\dfrac{64}{5}\ell n(4)-\dfrac{128}{25}+1$

$\displaystyle\sf\int_{1}^{4}x^{\frac{3}{2}}\ell n(x)~dx=\dfrac{320\ell n(4)-128+25}{25}\\\displaystyle\sf\int_1^4 x^{\frac{3}{2}}\ell n(x)~dx=\dfrac{320\ell n(4)-103}{25}\blue{\checkmark}$