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Making the u substitution \(\displaystyle u = \ln{\left(x \right)}\) gives \(\displaystyle du = \left(\frac{1}{x}\right)dx\). Solving for \(\displaystyle dx\) gives \(\displaystyle dx = xdu\). Substituting in the values of \(\displaystyle u\) and \(\displaystyle du\) gives \(\displaystyle \int \left(- \frac{5}{x \left(u^{2} + 1\right)}\right)\left(xdu\right)\). Simplifying gives the integral \(\displaystyle \int - \frac{5}{u^{2} + 1} du\). Integrating gives \(\displaystyle \int - \frac{5}{u^{2} + 1} du = - 5 \tan^{-1}{\left(u \right)}+C\). Substituting \(\displaystyle u\) back in gives the solution \(\displaystyle - 5 \tan^{-1}{\left(\ln{\left(x \right)} \right)}+C\).

Download \(\LaTeX\)

\begin{question}Find the indefinite integral of $\int \left(- \frac{5}{x \left(\ln{\left(x \right)}^{2} + 1\right)}\right)\, dx$. 
    \soln{9cm}{Making the u substitution $u = \ln{\left(x \right)}$ gives $du = \left(\frac{1}{x}\right)dx$. Solving for $dx$ gives $dx = xdu$. Substituting in the values of $u$ and $du$ gives $\int \left(- \frac{5}{x \left(u^{2} + 1\right)}\right)\left(xdu\right)$.  Simplifying gives the integral $\int - \frac{5}{u^{2} + 1} du$. Integrating gives  $\int - \frac{5}{u^{2} + 1} du = - 5 \tan^{-1}{\left(u \right)}+C$. Substituting $u$ back in gives the solution $- 5 \tan^{-1}{\left(\ln{\left(x \right)} \right)}+C$. }

\end{question}

Download Question and Solution Environment\(\LaTeX\)

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\begin{document}\begin{question}(10pts) The question goes here!
    \soln{9cm}{The solution goes here.}

\end{question}\end{document}

HTML for Canvas

<p> <p>Find the indefinite integral of  <img class="equation_image" title=" \displaystyle \int \left(- \frac{5}{x \left(\ln{\left(x \right)}^{2} + 1\right)}\right)\, dx " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cleft%28-%20%5Cfrac%7B5%7D%7Bx%20%5Cleft%28%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%5E%7B2%7D%20%2B%201%5Cright%29%7D%5Cright%29%5C%2C%20dx%20" alt="LaTeX:  \displaystyle \int \left(- \frac{5}{x \left(\ln{\left(x \right)}^{2} + 1\right)}\right)\, dx " data-equation-content=" \displaystyle \int \left(- \frac{5}{x \left(\ln{\left(x \right)}^{2} + 1\right)}\right)\, dx " /> . </p> </p>

HTML for Canvas

<p> <p>Making the u substitution  <img class="equation_image" title=" \displaystyle u = \ln{\left(x \right)} " src="/equation_images/%20%5Cdisplaystyle%20u%20%3D%20%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20" alt="LaTeX:  \displaystyle u = \ln{\left(x \right)} " data-equation-content=" \displaystyle u = \ln{\left(x \right)} " />  gives  <img class="equation_image" title=" \displaystyle du = \left(\frac{1}{x}\right)dx " src="/equation_images/%20%5Cdisplaystyle%20du%20%3D%20%5Cleft%28%5Cfrac%7B1%7D%7Bx%7D%5Cright%29dx%20" alt="LaTeX:  \displaystyle du = \left(\frac{1}{x}\right)dx " data-equation-content=" \displaystyle du = \left(\frac{1}{x}\right)dx " /> . Solving for  <img class="equation_image" title=" \displaystyle dx " src="/equation_images/%20%5Cdisplaystyle%20dx%20" alt="LaTeX:  \displaystyle dx " data-equation-content=" \displaystyle dx " />  gives  <img class="equation_image" title=" \displaystyle dx = xdu " src="/equation_images/%20%5Cdisplaystyle%20dx%20%3D%20xdu%20" alt="LaTeX:  \displaystyle dx = xdu " data-equation-content=" \displaystyle dx = xdu " /> . Substituting in the values of  <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX:  \displaystyle u " data-equation-content=" \displaystyle u " />  and  <img class="equation_image" title=" \displaystyle du " src="/equation_images/%20%5Cdisplaystyle%20du%20" alt="LaTeX:  \displaystyle du " data-equation-content=" \displaystyle du " />  gives  <img class="equation_image" title=" \displaystyle \int \left(- \frac{5}{x \left(u^{2} + 1\right)}\right)\left(xdu\right) " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20%5Cleft%28-%20%5Cfrac%7B5%7D%7Bx%20%5Cleft%28u%5E%7B2%7D%20%2B%201%5Cright%29%7D%5Cright%29%5Cleft%28xdu%5Cright%29%20" alt="LaTeX:  \displaystyle \int \left(- \frac{5}{x \left(u^{2} + 1\right)}\right)\left(xdu\right) " data-equation-content=" \displaystyle \int \left(- \frac{5}{x \left(u^{2} + 1\right)}\right)\left(xdu\right) " /> .  Simplifying gives the integral  <img class="equation_image" title=" \displaystyle \int - \frac{5}{u^{2} + 1} du " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20-%20%5Cfrac%7B5%7D%7Bu%5E%7B2%7D%20%2B%201%7D%20du%20" alt="LaTeX:  \displaystyle \int - \frac{5}{u^{2} + 1} du " data-equation-content=" \displaystyle \int - \frac{5}{u^{2} + 1} du " /> . Integrating gives   <img class="equation_image" title=" \displaystyle \int - \frac{5}{u^{2} + 1} du = - 5 \tan^{-1}{\left(u \right)}+C " src="/equation_images/%20%5Cdisplaystyle%20%5Cint%20-%20%5Cfrac%7B5%7D%7Bu%5E%7B2%7D%20%2B%201%7D%20du%20%3D%20-%205%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28u%20%5Cright%29%7D%2BC%20" alt="LaTeX:  \displaystyle \int - \frac{5}{u^{2} + 1} du = - 5 \tan^{-1}{\left(u \right)}+C " data-equation-content=" \displaystyle \int - \frac{5}{u^{2} + 1} du = - 5 \tan^{-1}{\left(u \right)}+C " /> . Substituting  <img class="equation_image" title=" \displaystyle u " src="/equation_images/%20%5Cdisplaystyle%20u%20" alt="LaTeX:  \displaystyle u " data-equation-content=" \displaystyle u " />  back in gives the solution  <img class="equation_image" title=" \displaystyle - 5 \tan^{-1}{\left(\ln{\left(x \right)} \right)}+C " src="/equation_images/%20%5Cdisplaystyle%20-%205%20%5Ctan%5E%7B-1%7D%7B%5Cleft%28%5Cln%7B%5Cleft%28x%20%5Cright%29%7D%20%5Cright%29%7D%2BC%20" alt="LaTeX:  \displaystyle - 5 \tan^{-1}{\left(\ln{\left(x \right)} \right)}+C " data-equation-content=" \displaystyle - 5 \tan^{-1}{\left(\ln{\left(x \right)} \right)}+C " /> . </p> </p>