How Can I Use Substitution to Solve This Integral?

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In summary, the formula for finding the integral of 2x/(x^2+9) dx is ∫(2x)/(x^2+9) dx = ln|x^2+9| + C. To solve this integral, use the substitution method and then substitute back in after integration. It can also be solved using partial fractions. The indefinite integral is the same as the original formula. The graph of 2x/(x^2+9) dx is a hyperbola with asymptotes at x = -3 and x = 3 and approaching the x-axis as x approaches infinity in either direction.
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karush
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$$\int\frac{2x}{x^2+9}\ \text{dx}$$
I thot I could use
$$u={x}^{2}+9$$
But counldn't go thru with it
 
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karush said:
$$\int\frac{2x}{x^2+9}\ \text{dx}$$
I thot I could use
$$u={x}^{2}+9$$
But counldn't go thru with it

(Wave)

$$u=x^2+9 \Rightarrow du=2x dx $$

$$\int\frac{2x}{x^2+9} dx=\int \frac{du}{u}=\ln |u|+c=\ln |x^2+9|+c=\ln(x^2+9)+c$$
 

FAQ: How Can I Use Substitution to Solve This Integral?

1. What is the formula for finding the integral of 2x/(x^2+9) dx?

The formula for finding the integral of 2x/(x^2+9) dx is ∫(2x)/(x^2+9) dx = ln|x^2+9| + C, where C is the constant of integration.

2. How do you solve the integral of 2x/(x^2+9) dx?

To solve the integral of 2x/(x^2+9) dx, use the substitution method by letting u = x^2+9. Then, find du/dx and substitute it into the integral. The integral then becomes ∫(2x)/(u) du. After integration, substitute back u = x^2+9 and add the constant of integration.

3. Can the integral of 2x/(x^2+9) dx be solved using partial fractions?

Yes, the integral of 2x/(x^2+9) dx can be solved using partial fractions. The fraction can be rewritten as 2x/(x^2+9) = A/(x+3) + B/(x-3), where A and B are constants. Then, solve for A and B and integrate each term separately.

4. What is the indefinite integral of 2x/(x^2+9) dx?

The indefinite integral of 2x/(x^2+9) dx is ∫(2x)/(x^2+9) dx = ln|x^2+9| + C, where C is the constant of integration.

5. How does the graph of 2x/(x^2+9) dx look like?

The graph of 2x/(x^2+9) dx is a hyperbola with two branches. The vertical asymptotes are x = -3 and x = 3, and the horizontal asymptote is y = 0. The graph approaches the x-axis as x approaches positive or negative infinity.

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