Given:
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx[/tex]
Required:
We need to evaluate the given integral.
Explanation:
[tex]Use\text{ 16=4}^2.[/tex]
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=\int_0^3\frac{x}{\sqrt{x^2+4^2}}dx[/tex][tex]Let\text{ u=x and dv=}\frac{1}{\sqrt{x^2+4^2}}dx.[/tex]
[tex]d\text{u=dx and }\int\text{dv=}\int\frac{1}{\sqrt{x^2+4^2}}dx.[/tex]
[tex]v=In\lvert{\frac{x+\sqrt{x^2+4^2}}{4}}\rvert[/tex][tex]Use\text{ u}\cdot dv=uv-\int vdu.[/tex]
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=xIn\lvert{\frac{x+\sqrt{x^2+4^2}}{4}}\rvert-\int_0^3In\lvert{\frac{x+\sqrt{x^2+4^2}}{4}}\rvert dx[/tex]
We know that
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=In\lvert\sqrt{x^2+16}+x|+C[/tex]
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=In|8|-In|4|[/tex]
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=1[/tex]
Final answer:
[tex]\int_0^3\frac{x}{\sqrt{x^2+16}}dx=1[/tex]
