Q. \[ \int \frac{1}{1+x^{2}}\,dx \]

Q: What is the antiderivative of \frac{1}{1+x^2} ?

\int \frac{1}{1+x^2}\,dx = \arctan(x) + C .

Q: Why does \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2} ?

Use the derivative \frac{d}{dx}\arctan(x)=\frac{1}{1+x^2} .

Q: Compute \int \frac{x}{1+x^2}\,dx .

Let u=1+x^2, then du=2x\,dx. Result: \frac{1}{2}\ln(1+x^2)+C .

Q: Find \int \frac{1}{1+(2x)^2}\,dx .

\int \frac{1}{1+4x^2}\,dx = \frac{1}{2}\arctan(2x)+C .

Q: Evaluate \int_{0}^{1} \frac{1}{1+x^2}\,dx .

\arctan(1)-\arctan(0)=\frac{\pi}{4} .

Answer

We want an antiderivative of \( \frac{1}{1+x^2} \). Recall the standard result:

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

Differentiate to check: \( \frac{d}{dx}\big(\arctan(x)\big)=\frac{1}{1+x^2} \).

\(\boxed{\arctan(x) + C}\)

Detailed Explanation

We want to find an antiderivative of the function

\[ \frac{1}{1+x^2}. \]

That means we want a function \(F(x)\) such that

\[ F'(x)=\frac{1}{1+x^2}. \]

Step 1: Recognize a standard derivative form.

A very common calculus result is that the derivative of the arctangent function is

\[ \frac{d}{dx}\left(\arctan(x)\right)=\frac{1}{1+x^2}. \]

Step 2: Match the given function to the known derivative.

Our integrand is exactly \(\frac{1}{1+x^2}\), which matches the derivative above.

Step 3: Write the antiderivative.

Since the derivative of \(\arctan(x)\) is \(\frac{1}{1+x^2}\), an antiderivative of \(\frac{1}{1+x^2}\) is

\[ \arctan(x)+C. \]

Final answer:

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

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Calculus FAQ

What is the antiderivative of \( \frac{1}{1+x^2} \)?

\( \int \frac{1}{1+x^2}\,dx = \arctan(x) + C \).

Why does \( \frac{d}{dx}\arctan(x) = \frac{1}{1+x^2} \)?

Use the derivative \( \frac{d}{dx}\arctan(x)=\frac{1}{1+x^2} \).

Compute \( \int \frac{1}{a^2+x^2}\,dx \) for \(a>0\).

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

Compute \( \int \frac{x}{1+x^2}\,dx \).

Let \(u=1+x^2\), then \(du=2x\,dx\). Result: \( \frac{1}{2}\ln(1+x^2)+C \).

Find \( \int \frac{1}{1+(2x)^2}\,dx \).

\( \int \frac{1}{1+4x^2}\,dx = \frac{1}{2}\arctan(2x)+C \).

Evaluate \( \int_{0}^{1} \frac{1}{1+x^2}\,dx \).

\( \arctan(1)-\arctan(0)=\frac{\pi}{4} \).

Solve the integral step by step.
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