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4vir4ik [10]
3 years ago
14

Find the general indefinite integral. (Use C for the constant of integration. Remember to use absolute values where appropriate.

) integral 9 + Squareroot x + x/x dx x + Squareroot x + 9 log (x) +
Mathematics
1 answer:
Maru [420]3 years ago
4 0

Answer:

9\text{ln}|x|+2\sqrt{x}+x+C

Step-by-step explanation:

We have been an integral \int \frac{9+\sqrt{x}+x}{x}dx. We are asked to find the general solution for the given indefinite integral.

We can rewrite our given integral as:

\int \frac{9}{x}+\frac{\sqrt{x}}{x}+\frac{x}{x}dx

\int \frac{9}{x}+\frac{1}{\sqrt{x}}+1dx

Now, we will apply the sum rule of integrals as:

\int \frac{9}{x}dx+\int \frac{1}{\sqrt{x}}dx+\int 1dx

9\int \frac{1}{x}dx+\int x^{-\frac{1}{2}}dx+\int 1dx

Using common integral \int \frac{1}{x}dx=\text{ln}|x|, we will get:

9\text{ln}|x|+\int x^{-\frac{1}{2}}dx+\int 1dx

Now, we will use power rule of integrals as:

9\text{ln}|x|+\frac{x^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+\int 1dx

9\text{ln}|x|+\frac{x^{\frac{1}{2}}}{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2x^{\frac{1}{2}}+\int 1dx

9\text{ln}|x|+2\sqrt{x}+\int 1dx

We know that integral of a constant is equal to constant times x, so integral of 1 would be x.

9\text{ln}|x|+2\sqrt{x}+x+C

Therefore, our required integral would be 9\text{ln}|x|+2\sqrt{x}+x+C.

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<u>Answer:</u>

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