Q. \(\displaystyle \int \frac{1}{x^2+4}\,dx\)

Q: What substitution matches x^2+4 to an arctangent form?

Use x=2u, so dx=2du, giving \int \frac{1}{x^2+4}\,dx=\frac{1}{2}\int \frac{1}{u^2+1}\,du.

Q: Why does a^2+x^2 lead to an \arctan result?

Because \displaystyle \int \frac{1}{x^2+a^2}\,dx=\frac{1}{a}\arctan\!\left(\frac{x}{a}\right)+C, for a>0.

Q: What is \int \frac{1}{4+x^2}\,dx compared to the original?

It is the same integral since 4+x^2=x^2+4, so the result is \displaystyle \frac{1}{2}\arctan\!\left(\frac{x}{2}\right)+C.

Q: What is \int \frac{1}{x^2+4} \, dx using a trig substitution?

Let x=2\tan\theta. Then dx=2\sec^2\theta\,d\theta and x^2+4=4\sec^2\theta, so the integral becomes \frac{1}{2}\int d\theta=\frac{1}{2}\theta+C=\frac{1}{2}\arctan(x/2)+C.

Answer

Compute

\[
\int \frac{1}{x^2+4}\,dx.
\]

Use the form

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

Here \(a=2\), so

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

Detailed Explanation

We want to compute the indefinite integral

\[
\int \frac{1}{x^2+4}\,dx
\]

Step 1: Recognize the standard form

A very common integral to know is:

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

Here, we want to match the denominator \(x^2+4\) to the form \(u^2+a^2\).

Step 2: Match parameters

Compare:

\(x^2+4 = x^2 + 2^2\)
So we have \(u=x\) and \(a=2\)

Step 3: Apply the formula

Using \(a=2\) gives:

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

Step 4: State the final answer

\[
\int \frac{1}{x^2+4}\,dx=\frac{1}{2}\arctan\left(\frac{x}{2}\right)+C
\]

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Calculus FAQ

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

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

What substitution matches \(x^2+4\) to an arctangent form?

Use \(x=2u\), so \(dx=2du\), giving \(\int \frac{1}{x^2+4}\,dx=\frac{1}{2}\int \frac{1}{u^2+1}\,du\).

Why does \(a^2+x^2\) lead to an \(\arctan\) result?

Because \(\displaystyle \int \frac{1}{x^2+a^2}\,dx=\frac{1}{a}\arctan\!\left(\frac{x}{a}\right)+C\), for \(a>0\).

Can you rewrite the answer using \( \arctan \) with different scaling?

\(\displaystyle \frac{1}{2}\arctan\!\left(\frac{x}{2}\right)+C\) is equivalent to \(\displaystyle \frac{1}{\sqrt{4}}\arctan\!\left(\frac{x}{\sqrt{4}}\right)+C\).

What is \( \int \frac{1}{4+x^2}\,dx \) compared to the original?

It is the same integral since \(4+x^2=x^2+4\), so the result is \(\displaystyle \frac{1}{2}\arctan\!\left(\frac{x}{2}\right)+C\).

What is the definite integral from \(0\) to \(2\)?

\(\displaystyle \int_{0}^{2}\frac{1}{x^2+4}\,dx=\left.\frac{1}{2}\arctan\!\left(\frac{x}{2}\right)\right|_{0}^{2}=\frac{1}{2}\left(\frac{\pi}{4}-0\right)=\frac{\pi}{8}\).

What is \( \int \frac{1}{x^2+4} \, dx \) using a trig substitution?

Let \(x=2\tan\theta\). Then \(dx=2\sec^2\theta\,d\theta\) and \(x^2+4=4\sec^2\theta\), so the integral becomes \(\frac{1}{2}\int d\theta=\frac{1}{2}\theta+C=\frac{1}{2}\arctan(x/2)+C\).

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