Respuesta :
Answer:
[tex]9\text{ln}|x|+2\sqrt{x}+x+C[/tex]
Step-by-step explanation:
We have been an integral [tex]\int \frac{9+\sqrt{x}+x}{x}dx[/tex]. We are asked to find the general solution for the given indefinite integral.
We can rewrite our given integral as:
[tex]\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx[/tex]
[tex]\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx[/tex]
Now, we will apply the sum rule of integrals as:
[tex]\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx[/tex]
[tex]9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx[/tex]
Using common integral [tex]\int \frac{1}{x}dx=\text{ln}|x|[/tex], we will get:
[tex]9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx[/tex]
Now, we will use power rule of integrals as:
[tex]9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx[/tex]
[tex]9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx[/tex]
[tex]9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx[/tex]
[tex]9\text{ln}|x|+2\sqrt{x}+\int 1dx[/tex]
We know that integral of a constant is equal to constant times x, so integral of 1 would be x.
[tex]9\text{ln}|x|+2\sqrt{x}+x+C[/tex]
Therefore, our required integral would be [tex]9\text{ln}|x|+2\sqrt{x}+x+C[/tex].
