Respuesta :

[tex]\frac{d}{dx}\int x^3\ln x^2dx[/tex]

METHOD : We will use integration by part

[tex]\begin{gathered} \int udv=uv-\int vdu \\ \end{gathered}[/tex]

I will modify the question by putting "1" before the In(x^2) function. See modified question in equation tab below

[tex]\begin{gathered} \int1\ln(x^2)dx \\ u=\ln(x^2) \\ \frac{du}{dx}=\frac{2x}{x^2}=\frac{2}{x} \\ \\ dv=1 \\ v=\int1dx \\ v=x \\ \end{gathered}[/tex][tex]\begin{gathered} \int1\ln(x^2)=x\ln x^2-\int x\frac{2dx}{x} \\ \\ =x\ln x^2-\int2dx \\ \\ =x\ln x^2-2x \end{gathered}[/tex]

[tex]\begin{gathered} \int_2^{x^3}[x\ln x^2-2x]dx \\ \\ [x^3\ln(x^3)^2-2(x^{`3})]-[2\ln2^2-2(2)] \\ x^3\ln x^6-2x^3-2\ln4-4 \end{gathered}[/tex]

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