Respuesta :
Answer:
The given integral is convergent and its value is [tex]\frac{1}{3}[/tex].
Step-by-step explanation:
We have been given a definite integral [tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}[/tex]. We are asked to determine whether the given integral converges or diverges.
We will use u-substitution to solve our given integral as:
Let [tex]u=3x+1[/tex].
[tex]\frac{du}{dx}=3[/tex]
[tex]dx=\frac{1}{3}du[/tex]
[tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}=\int _0^{\infty }\:\frac{1}{\left(u\right)^2}*\frac{1}{3}du[/tex]
[tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}=\frac{1}{3}\int _0^{\infty }\:\frac{1}{\left(u\right)^2}du[/tex]
[tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}=\frac{1}{3}\int _0^{\infty }\:u^{-2}du[/tex]
[tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}=\frac{1}{3}*\frac{u^{-2+1}}{-2+1}[/tex]
[tex]\int _0^{\infty }\:\frac{dx}{\left(3x+1\right)^2}=\frac{1}{3}*\frac{u^{-1}}{-1}=-\frac{1}{3u}=-\frac{1}{3(3x+1)}[/tex]
Now, we will compute the boundaries as:
[tex]-\frac{1}{3(3(\infty)+1)}=-\frac{1}{\infty}=0[/tex]
[tex]-\frac{1}{3(3(0)+1)}=-\frac{1}{3(1)}=-\frac{1}{3}[/tex]
[tex]0-(-\frac{1}{3})=\frac{1}{3}[/tex]
Therefore, the given integral is convergent and its value is [tex]\frac{1}{3}[/tex].
