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Given:


\int_0^3(x)/(√(x^2+16))dx

Required:

We need to evaluate the given integral.

Step-by-step explanation:


Use\text{ 16=4}^2.


\int_0^3(x)/(√(x^2+16))dx=\int_0^3(x)/(√(x^2+4^2))dx
Let\text{ u=x and dv=}(1)/(√(x^2+4^2))dx.


d\text{u=dx and }\int\text{dv=}\int(1)/(√(x^2+4^2))dx.


v=In\lvert{(x+√(x^2+4^2))/(4)}\rvert
Use\text{ u}\cdot dv=uv-\int vdu.


\int_0^3(x)/(√(x^2+16))dx=xIn\lvert{(x+√(x^2+4^2))/(4)}\rvert-\int_0^3In\lvert{(x+√(x^2+4^2))/(4)}\rvert dx

We know that


\int_0^3(x)/(√(x^2+16))dx=In\lvert√(x^2+16)+x|+C
\int_0^3(x)/(√(x^2+16))dx=In|8|-In|4|
\int_0^3(x)/(√(x^2+16))dx=1

Final answer:


\int_0^3(x)/(√(x^2+16))dx=1

User Khanh Tran
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