Question

Ex 4.8 14) integrate 3^(2x-1)

Answer 1

Take the integral:

$\large\begin{array}{l} \mathsf{\displaystyle\int\!3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\int\!\frac{1}{2}\cdot 2\cdot 3^{2x-1}\,dx}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^{2x-1}\cdot 2\,dx\qquad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{2x-1=u\quad\Rightarrow\quad 2\,dx=du}\\\\\\ \textsf{so (i) becomes}\\\\ =\mathsf{\displaystyle\frac{1}{2}\int\!3^u\,du}\\\\ =\mathsf{\dfrac{1}{2}\cdot \dfrac{1}{\ell n\,3}\,3^u+C}\\\\ =\mathsf{\dfrac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}$


$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{\displaystyle\int\!3^{2x-1}\,dx=\frac{1}{2\,\ell n\,3}\,3^{2x-1}+C} \end{array}}\qquad\checkmark \end{array}$


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$\large\textsf{I hope it helps. :-)}$


Tags: integrate indefinite integral substitution exponential base logarithm log ln composite integral calculus