Answer to Question #176988 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

1. ∫(dx)/(x²+4x+5)

2. ∫(dt)/(t²-t+2)

Answer 1

$1)x^2 + 4x +5 = (x+2)^2 +1$

So we have that

$\int \dfrac{dx}{x^2 +4x+5} = \int \dfrac{dx}{(x+2)^2 +1} = \arctan (x+2) + c$

where c is a constant

2) $t^2-t+2 = (t-\frac{1}{2})^2 + \frac{7}{4} = (t-\frac{1}{2})^2 + (\frac{\sqrt7}{2})^2$

So we have that

$\int \dfrac{dt}{t^2-t+2} = \int \dfrac{dt}{(t-\frac{1}{2})^2 + (\frac{\sqrt7}{2})^2} = \frac{1}{\frac{\sqrt7}{2}} \arctan(\frac{t- \frac{1}{2}}{\frac{\sqrt7}{2}}) + b$ = $\frac{2}{\sqrt7} \arctan(\frac{2t-1}{\sqrt7}) + b$

where b is a constant