Question

Determinar as seguintes integrais usando a técnica de integração por substituição:

Em anexo!

Answer 1

a)
Seja x²-1 = u, logo 2x dx = 1du, então
$dx = \frac{du}{2x}$
Substitua estes valores na integral:

$\int( \frac{x}{ {x}^{2} - 1} dx)$
=
$\int( \frac{x}{ u} \frac{du}{2x} )$
=
$\int( \frac{1}{ u} \frac{du}{2} )$
=
$\frac{1}{2} \int( \frac{1}{ u} du )$
=
1/2 • ln |u| + C
= ln |x²-1| • 1/2 + C

b)
integral de (x^3 - x^2 + 1)/(x + 1) dx

Faça a divisão polinomial:

= integral (x^2 - 2 x - 1/(x + 1) + 2) dx

= - int 1/(x + 1) dx + int (x^2) dx - 2 int x dx + 2 int (1) dx

Para integrar 1/(x+1), substitua u= x+1:

= - integral 1/u du + integral x^2 dx - 2 integral x dx + 2 integral de 1 dx
=
-ln|u|+ (x^3)/3 - x² + 2x

Substitua 'de volta' x = u+1
= x^3/3 - x^2 + 2 x - ln|x + 1| + C

Answer 2

$a)\ \int\ { \dfrac{x}{x^2-1} } \, dx \\\\\\u=x^2-1\\du=2xdx\\\\xdx= \dfrac{1}{2}du\\\\\\ \int\ { \dfrac{1}{2}\cdot \dfrac{1}{u} } \, du\\\\\\ \dfrac{1}{2}\cdot \int\ { \dfrac{1}{u} } \, du \\\\\\ \dfrac{1}{2} \cdot{ln|u|}+C\\\\\\ \dfrac{1}{2}\cdot{{ln|x^2-1]+C$


$b)\ \int\limits^2_1 { \dfrac{x^3-x^2+1}{x+1}} \, dx\\\\\\divis\~ao\ de\ polin\^omios\ (segue\ em\ anexo)$

$\int\limits^2_1 { \bigg[\dfrac{(x^2-2x+2)\cdot(x+1)-1}{x+1}\bigg] } \, dx\\\\\\ \int\limits^2_1 { \dfrac{(x^2-2x+2)\cdot(x+1)}{x+1}} \, dx- \int\limits^2_1 { \dfrac{1}{x+1} \, dx$

$u=x+1\\du=dx\\\\\\ \int\limits^2_1 {(x^2-2x+2)} \, dx- \int\limits^2_1 { \dfrac{1}{u} } \, du\\\\\\ \dfrac{x^3}{3}- \dfrac{2x^2}{2}+2x-ln|x+1|\bigg|_1^2\\\\\\ \dfrac{2^3}{3}-(2)^2+2\cdot(2)-ln|2+1|-\bigg[ \dfrac{1}{3}^3-(1)^2+2\cdot(1)-ln|1+1|\bigg]\\\\\\ \dfrac{8}{3}-4+4-ln|3|- \dfrac{1}{3}+1-2+ln|2|\\\\\\ \dfrac{10}{3}-ln|3|-ln|2|\\\\\\ \dfrac{10}{3}- \dfrac{ln|3|}{ln|2|}$