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Evgesh-ka [11]
3 years ago
12

Integral of 11 ln(cube root(x)) dx?

Mathematics
2 answers:
Darina [25.2K]3 years ago
5 0
If you're using the app, try seeing this answer through your browser:  brainly.com/question/2742530

_______________


Evaluate the indefinite integral:

\mathsf{\displaystyle\int 11\,\ell n(^3\hspace{-5}\sqrt{x})\,dx}\\\\\\ \mathsf{\displaystyle\int\!\ell n(^3\hspace{-5}\sqrt{x})\cdot 11\,dx\qquad\quad(i)}


Integration by parts:

\begin{array}{lcl}
\mathsf{u=\,\ell n(^3\hspace{-5}\sqrt{x})}&\quad\Rightarrow\quad&\mathsf{du=\dfrac{1}{^3\hspace{-5}\sqrt{x}}\cdot \dfrac{d}{dx}(^3\hspace{-5}\sqrt{x})\,dx}\\\\
&&\mathsf{du=x^{-1/3}\cdot \dfrac{1}{3}\,x^{(1/3)-1}\,dx}\\\\
&&\mathsf{du=\dfrac{1}{3}\,x^{-1/3}\cdot x^{-2/3}\,dx}\\\\
&&\mathsf{du=\dfrac{1}{3}\,x^{-1}\,dx}
\\\\\\
\mathsf{dv=11\,dx}&\quad\Leftarrow\quad&\mathsf{v=11x}
\end{array}


\mathsf{\displaystyle\int\! u\,dv=u\cdot v-\int\!v\,du}\\\\\\
\mathsf{\displaystyle\int\!\ell n(^3\hspace{-5}\sqrt{x})\cdot 11\,dx=\ell n(^3\hspace{-5}\sqrt{x})\cdot 11x-\int\!11x\cdot \frac{1}{3}\,x^{-1}\,dx}\\\\\\
\boxed{\begin{array}{c}\mathsf{\displaystyle\int\!11\,\ell n(^3\hspace{-5}\sqrt{x})\,dx=11x\,\ell n(^3\hspace{-5}\sqrt{x})-\frac{11}{3}x+C}\end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)

Leno4ka [110]3 years ago
3 0
\bf \int [11\ ln\left( \sqrt[3]{x} \right)]dx\qquad 
\begin{cases}
ln\left( \sqrt[3]{x} \right)\implies ln\left( x^{\frac{1}{3}} \right)
\\ \quad \\
\frac{1}{3}ln(x)
\end{cases}\qquad thus
\\ \quad \\\\ \quad \\
\int [11\ ln\left( \sqrt[3]{x} \right)]dx\implies \int[11\cdot \frac{1}{3}ln(x)]dx\impliedby \textit{scalars to the left}
\\ \quad \\
11\cdot \frac{1}{3}\int[ln(x)]dx\implies \cfrac{11}{3}\int[ln(x)]dx

and I'm pretty sure you know what that is
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