Question

Integral Tables Antiderivatives of various generic functions can be found as formulas in integral tables. See if you can derive the formulas that would appear in an integral table for the fol- lowing functions. (Here, \(a\) is an arbitrary constant.) See below. (a) \(\int \frac{d x}{a^{2}+x^{2}} \quad\) (b) \(\int \frac{d x}{a^{2}-x^{2}} \quad\) (c) \(\int \frac{d x}{(a+x)^{2}}\)

Short answer

The antiderivatives of the given functions are: (a) \(\frac{1}{a} \arctan(\frac{x}{a}) + C\), (b) \(\frac{1}{2a} \ln |\frac{a + x}{a - x}| + C\), and (c) \(-\frac{1}{a+x} + C\).

Step-by-step solution

Solve Integral (a)

The integral \(\int \frac{1}{a^{2}+x^{2}} dx\) can be solved using the basic antiderivative formula for \(\int \frac{1}{a^2 + x^2} dx = \frac{1}{a} \arctan(\frac{x}{a}) + C\) where C is the constant of integration. Therefore, the antiderivative of \(\frac{1}{a^{2}+x^{2}}\) is \(\frac{1}{a} \arctan(\frac{x}{a}) + C\).

Solve Integral (b)

The integral \(\int \frac{1}{a^{2}-x^{2}} dx\) can be solved using the basic anti-derivative formula for \(\int \frac{1}{a^{2} - x^2} dx = \frac{1}{2a} \ln |\frac{a + x}{a - x}| + C\), with \(a > 0\). So, the antiderivative of the given function is \(\frac{1}{2a} \ln |\frac{a + x}{a - x}| + C\).

Solve Integral (c)

For \(\int \frac{1}{(a+x)^{2}} dx\), note that it can be rewritten as \(\int \frac{1}{u^2} du\), where \(u = a + x\). The antiderivative of \(\frac{1}{u^2}\) is \(-\frac{1}{u}\), hence the antiderivative of the given function is \(-\frac{1}{a+x} + C\).