Question

find the antiderivative for each function when c equals 0. do as 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 $\frac{1}{3}\arctan(3x)$.

Explanation:

Step1: Recall the antiderivative formula for \(\frac{1}{\sqrt{1 - x^{2}}}\)

The antiderivative of \(\frac{1}{\sqrt{1 - x^{2}}}\) is \(\arcsin(x)\). Using the constant - multiple rule \(\int kf(x)dx=k\int f(x)dx\) (\(k = 3\)), so \(\int\frac{3}{\sqrt{1 - x^{2}}}dx=3\int\frac{1}{\sqrt{1 - x^{2}}}dx\).

Step2: Recall the antiderivative formula for \(\frac{1}{x^{2}+1}\)

The antiderivative of \(\frac{1}{x^{2}+1}\) is \(\arctan(x)\). Using the constant - multiple rule (\(k=\frac{1}{4}\)), so \(\int\frac{1}{4(x^{2}+1)}dx=\frac{1}{4}\int\frac{1}{x^{2}+1}dx\).

Step3: Use substitution for \(\int\frac{1}{1 + 9x^{2}}dx\)

Let \(u = 3x\), then \(du=3dx\) and \(dx=\frac{1}{3}du\). So \(\int\frac{1}{1 + 9x^{2}}dx=\frac{1}{3}\int\frac{1}{1+u^{2}}du\). Since \(\int\frac{1}{1 + u^{2}}du=\arctan(u)+C\), substituting back \(u = 3x\) gives \(\frac{1}{3}\arctan(3x)\).

Step4: Differentiate to check

• For \(y = 3\arcsin(x)\), using the formula \(\frac{d}{dx}(\arcsin(x))=\frac{1}{\sqrt{1 - x^{2}}}\), then \(y^\prime=3\times\frac{1}{\sqrt{1 - x^{2}}}\).
• For \(y=\frac{1}{4}\arctan(x)\), using the formula \(\frac{d}{dx}(\arctan(x))=\frac{1}{x^{2}+1}\), then \(y^\prime=\frac{1}{4}\times\frac{1}{x^{2}+1}\).
• For \(y=\frac{1}{3}\arctan(3x)\), using the chain rule \(\frac{d}{dx}(f(g(x)))=f^\prime(g(x))\cdot g^\prime(x)\). Since \(f(u)=\arctan(u)\), \(f^\prime(u)=\frac{1}{1 + u^{2}}\) and \(g(x)=3x\), \(g^\prime(x)=3\). Then \(y^\prime=\frac{1}{3}\times\frac{3}{1+(3x)^{2}}=\frac{1}{1 + 9x^{2}}\).

Answer:

a. \(3\arcsin(x)\)
b. \(\frac{1}{4}\arctan(x)\)
c. \(\frac{1}{3}\arctan(3x)\)