Question

How do I use substitution to integrate x*sqrt 3x^2+4?

Answer 1

Finding the indefinite integral by substitution:

$\large\begin{array}{l} \mathsf{\displaystyle\int x\sqrt{3x^2+4}\,dx}\\\\ =\mathsf{\displaystyle\int \frac{1}{6}\cdot 6x\sqrt{3x^2+4}\,dx}\\\\ =\mathsf{\displaystyle \frac{1}{6}\int\sqrt{3x^2+4}\cdot 6x\,dx\qquad\quad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Substitute}\\\\ \mathsf{3x^2+4=u~~\Rightarrow~~6x\,dx=du}\\\\\\ \textsf{then (i) becomes}\\\\ =\mathsf{\displaystyle \frac{1}{6}\int\sqrt{u}\,du}\\\\ =\mathsf{\displaystyle \frac{1}{6}\int u^\frac{1}{2}\,du}\\\\ =\mathsf{\dfrac{1}{6}\cdot \dfrac{u^{\frac{1}{2}+1}}{\frac{1}{2}+1}+C} \end{array}$

$\large\begin{array}{l} =\mathsf{\dfrac{1}{6}\cdot \dfrac{~u^{\frac{3}{2}}~}{\frac{3}{2}}+C}\\\\ =\mathsf{\dfrac{1}{6}\cdot \dfrac{2}{3}\,u^{\frac{3}{2}}+C}\\\\ =\mathsf{\dfrac{1\cdot 2}{6\cdot 3}\,u^{\frac{3}{2}}+C}\\\\ =\mathsf{\dfrac{1\cdot \diagup\!\!\!\! 2}{\diagup\!\!\!\! 2\cdot 3\cdot 3}\,u^{\frac{3}{2}}+C}\\\\ =\mathsf{\dfrac{1}{9}\,u^{\frac{3}{2}}+C} \end{array}$


$\large\begin{array}{l} \textsf{Substitute back for }\mathsf{u=3x^2+4:}\\\\ =\mathsf{\dfrac{1}{9}\,(3x^2+4)^{\frac{3}{2}}+C}\\\\\\ \therefore~~\boxed{\begin{array}{c} \mathsf{\displaystyle\int x\sqrt{3x^2+4}\,dx=\frac{1}{9}\,(3x^2+4)^{\frac{3}{2}}+C} \end{array}}\qquad\checkmark \end{array}$


If you're having problems understanding this answer, try seeing it through your browser: brainly.com/question/2097209


$\large\textsf{I hope it helps.}$


Tags: definite integral integrate limits function irrational square root sqrt composite substitution integral calculus