Integrals - Integrals of some more types

Evaluate $\int \sqrt{x^2 + 4x + 1} \, dx$.

First, complete the square for the expression $x^2 + 4x + 1$: $x^2 + 4x + 1 = (x^2 + 4x + 4) - 4 + 1 = (x + 2)^2 - 3$ So, the integral becomes: $I = \int \sqrt{(x + 2)^2 - (\sqrt{3})^2} \, dx$ Using the formula $\int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2}\log|x + \sqrt{x^2 - a^2}| + C$, where $X = x + 2$ and $a = \sqrt{3}$: $I = \frac{x+2}{2}\sqrt{(x+2)^2 - 3} - \frac{3}{2}\log|(x+2) + \sqrt{(x+2)^2 - 3}| + C$ $I = \frac{x+2}{2}\sqrt{x^2 + 4x + 1} - \frac{3}{2}\log|x + 2 + \sqrt{x^2 + 4x + 1}| + C$

We transform the quadratic into the standard form $\sqrt{X^2 - a^2}$ by completing the square and then apply the specific logarithmic formula.

Evaluate $\int \sqrt{1 - 4x - x^2} \, dx$.

Complete the square for $1 - 4x - x^2$: $1 - (x^2 + 4x) = 1 - (x^2 + 4x + 4 - 4) = 1 - (x + 2)^2 + 4 = 5 - (x + 2)^2$ The integral becomes: $I = \int \sqrt{(\sqrt{5})^2 - (x + 2)^2} \, dx$ Using the formula $\int \sqrt{a^2 - x^2} \, dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\left(\frac{x}{a}\right) + C$: $I = \frac{x+2}{2}\sqrt{5 - (x+2)^2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$ $I = \frac{x+2}{2}\sqrt{1 - 4x - x^2} + \frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right) + C$

Since the $x^2$ term is negative, the quadratic expression is rearranged into the form $a^2 - (x+h)^2$, leading to an inverse sine result.