Answer to Question #176991 in Calculus for Phyroe

Question

Integrals Giving Inverse Trigonometric Functions

1. ∫(x+1)/(x² +1) from 0 to 1

2. ∫(dz)/z(1 + ln² z) from 1 to e

Answer 1

1) $\int_{0}^1 \frac{x+1}{x^2 +1}dx$ = $\int_{0}^1 \frac{x}{x^2 +1} + \frac{1}{x^2 + 1}dx$

= $\int_{0}^1 \frac{x}{x^2 +1}dx + \int_{0}^1\frac{1}{x^2 + 1}dx$

= $\int_{0}^1 \frac{x}{x^2 +1}dx$ + tan^-1(x) $|_0^1$

let x² + 1 = u $\implies$ 2x$dx$ = $du$ & x = 0 then u = 1

$\implies$ x$dx$ = $\frac{du}{2}$ & x = 1 then u = 2

therefore given integral becomes,

= $\int_{1}^2 \frac{1}{2u}du$ + tan^-1(1) - tan^-1(0)

= $\int_{1}^2 \frac{1}{2u}du$ + $\frac{\pi}{4}$ - 0

= $\frac{1}{2}$ ln(u) $|_1^2$ + $\frac{\pi}{4}$

= $\frac{1}{2}$ ( ln(2) - ln(1) ) + $\frac{\pi}{4}$

= $\frac{1}{2}$ ln(2) + $\frac{\pi}{4}$

2) $\int_{1}^e \frac{1}{z(1 + ln^2 (z) }dz$ $\implies$ let $ln(z)$ = t & z = e then t = 1

$\implies$ $\frac{1}{z} dz = dt$ & z = 1 then t = 0

therefore integral becomes,

= $\int_{0}^1 \frac{1}{(1 + t^2) }dt$

= tan^-1(t) $|_0^1∣ ​$

= tan^-1(1) - tan^-1(0)

= $\frac{\pi}{4}$