Question

Evaluate the integral 8/ ((x+3) ln(x+3)) dx

Answer 1

Evaluate the following integral by a simple substitution:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!\frac{8}{(x+3)\,\ell n(x+3)}\,dx}\\\\ =\mathsf{\displaystyle 8\int\!\frac{1}{\ell n(x+3)}\cdot \frac{1}{x+3}\,dx\qquad(i)}\\\\ \end{array}$


$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{\ell n(x+3)=u\quad\Rightarrow\quad\dfrac{1}{x+3}\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle \!8\int\!\frac{1}{u}\,du}\\\\ =\mathsf{8\,\ell n|u|+C}\\\\ =\mathsf{8\,\ell n\big|\ell n(x+3)\big|+C} \end{array}$


$\large\boxed{\begin{array}{c}\mathsf{\displaystyle\int\!\frac{8}{(x+3)\,\ell n(x+3)}\,dx=8\,\ell n\big|\ell n(x+3)\big|+C} \end{array}}\qquad\checkmark$


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Tags: indefinite integral substitution logarithm rational fraction integral calculus