Question

find the antiderivative for each function when c equals 0. do as many as you can mentally. check your answers by differentiation.
a. $\frac{3}{\sqrt{1 - x^{2}}}$ b. $\frac{1}{4(x^{2}+1)}$ c. $\frac{1}{1 + 9x^{2}}$
a. the antiderivative of $\frac{3}{\sqrt{1 - x^{2}}}$ is $3\arcsin(x)$
b. the antiderivative of $\frac{1}{4(x^{2}+1)}$ is $\frac{1}{4}\arctan(x)$
c. the antiderivative of $\frac{1}{1 + 9x^{2}}$ is $\square$

Explanation:

Step1: Recall the antiderivative formula

The antiderivative formula for \(\frac{1}{1 + u^{2}}\) is \(\arctan(u)+C\). For the function \(\frac{1}{1 + 9x^{2}}\), we can rewrite it as \(\frac{1}{1+(3x)^{2}}\).
Let \(u = 3x\), then \(du=3dx\), and \(dx=\frac{1}{3}du\).

Step2: Integrate the function

\(\int\frac{1}{1 + 9x^{2}}dx=\int\frac{1}{1+(3x)^{2}}dx\). Substituting \(u = 3x\) and \(dx=\frac{1}{3}du\) into the integral, we get \(\frac{1}{3}\int\frac{1}{1 + u^{2}}du\).
Since \(\int\frac{1}{1 + u^{2}}du=\arctan(u)+C\), then \(\frac{1}{3}\int\frac{1}{1 + u^{2}}du=\frac{1}{3}\arctan(u)+C\).
Substituting back \(u = 3x\), we have \(\frac{1}{3}\arctan(3x)+C\). When \(C = 0\), the antiderivative is \(\frac{1}{3}\arctan(3x)\).

Step3: Check by differentiation

Differentiate \(y=\frac{1}{3}\arctan(3x)\) using the chain rule. The derivative of \(\arctan(u)\) with respect to \(u\) is \(\frac{1}{1+u^{2}}\). Let \(u = 3x\), then \(y^\prime=\frac{1}{3}\times\frac{3}{1+(3x)^{2}}=\frac{1}{1 + 9x^{2}}\)

Answer:

\(\frac{1}{3}\arctan(3x)\)