To solve the integral of \( \frac{1}{x^2 + 25} \), we can use a standard formula for integrals involving sums of squares:

\[
\int \frac{1}{x^2 + a^2} \, dx = \frac{1}{a} \arctan\left(\frac{x}{a}\right) + C
\]

In this case, \( a = 5 \), because \( 25 = 5^2 \). So, applying the formula:

\[
\int \frac{1}{x^2 + 25} \, dx = \frac{1}{5} \arctan\left(\frac{x}{5}\right) + C
\]

Thus, the integral is:

\[
\frac{1}{5} \arctan\left(\frac{x}{5}\right) + C
\]

Would you like a detailed breakdown of this solution, or do you have any questions?

Here are five related questions to expand on this:
1. How would you approach integrating \( \frac{1}{x^2 + b^2} \) for a different constant \( b \)?
2. What happens if the denominator was \( x^2 - 25 \) instead of \( x^2 + 25 \)?
3. Can you differentiate \( \frac{1}{5} \arctan\left(\frac{x}{5}\right) \) to verify the solution?
4. How does the integration change if you introduce an \( x \)-dependent term in the numerator?
5. What is the relationship between the arctangent function and trigonometric identities?

**Tip**: Always recognize sums of squares in the integrand as a cue to use inverse trigonometric functions for easier integration.

Solve the integral of 1/(x^2 + 25) with a step-by-step solution using inverse trigonometric functions. This calculus problem introduces key concepts such as the integral of sums of squares and the arctangent function. Suitable for college students studying calculus.

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