Find the Antiderivative f(x)=1/(x^2+81)

Solution:

Step 1:

Identify the antiderivative $F(x)$ by integrating the given function $f(x)$.

$F(x)=\int f(x)dx$

Step 2:

Write down the integral that needs to be solved.

$F(x)=\int \frac{1}{x^2+81}dx$

Step 3:

Transform the integral into a more solvable form.

Step 3.1:

Rearrange the terms in the denominator.

$\int \frac{1}{81+x^2}dx$

Step 3.2:

Express $81$ as a square of $9$.

$\int \frac{1}{9^2+x^2}dx$

Step 4:

Solve the integral using the formula for the arctangent function.

$\frac{1}{9}\arctan \left(\frac{x}{9}\right)+C$

Step 5:

Express the solution in its simplest form.

Step 5.1:

Factor out $\frac{1}{9}$ from the arctangent function.

$\frac{\arctan \left(\frac{x}{9}\right)}{9}+C$

Step 5.2:

Present the antiderivative in its final form.

$\frac{1}{9}\arctan \left(\frac{1}{9}x\right)+C$

Step 6:

Conclude with the antiderivative of the original function $f(x)=\frac{1}{x^2+81}$.

$F(x)=\frac{1}{9}\arctan \left(\frac{1}{9}x\right)+C$

Knowledge Notes:

To solve for the antiderivative of a function $f(x)=\frac{1}{x^2+a^2}$, where $a$ is a constant, we use the standard integral formula:

$$\int \frac{1}{x^2+a^2}dx=\frac{1}{a}\arctan \left(\frac{x}{a}\right)+C$$ where $C$ is the constant of integration. This formula is derived from the inverse trigonometric function arctangent (arctan or ${\tan }^{-1}$), which is the antiderivative of $\frac{1}{1+x^2}$.

In our specific problem, we have $a^2=81$, which means $a=9$. Therefore, we rewrite the integral in terms of $9^2$ to apply the formula correctly. After integrating, we simplify the expression by factoring out constants and combining terms to present the antiderivative in its simplest form.