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2 years ago

If x2 + mx + m is a perfect-square trinomial, which equation must be true?

Mathematics

2 answers:

Burka [1]2 years ago

8 0

Answer:

B. x2 + mx + m = (x + 1)2

Volgvan2 years ago

7 0

Your answer is x² + 2x + 1

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Maru [420]

Answer:

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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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2 years ago