Question

Calcule:

a) $\int\limits {x^2}* ln(x) \ * dx$

b) $\int\ \frac{(1- \sqrt{x} )^2}{2 \sqrt{x} } *dx$

c) $\int\ (x^2 + 2x - \sqrt{x} ) *dx$

Answer 1

a) Integral por partes:  $\int\limits {udv} \, =u.v- \int\limits{v} \, du$
Chamando $ln(x)$ de $u$ e $[tex]x^2$[/tex] de $dv$ fica: $du=- \frac{dx}{x};v= \frac{x^3}{3}$

$\frac{x^3.ln(x)}{3}- \int\limits { \frac{x^3}{3} } \, . \frac{dx}{x} => \frac{x^3ln(x)}{3}- \frac{1}{3} \int\limits{x^2} \, dx =>$$\frac{x^3ln(x)}{3}- \frac{1}{3} ( \frac{x^3}{3} ) => \frac{x^3ln(x)}{3}- \frac{x^3}{9}+C$

b) Por substituição:

$\int\limits { \frac{(1- \sqrt{x} )^2}{2 \sqrt{x} } } \, dx => \frac{1}{2} \int\limits { \frac{(1- \sqrt{x} )^2}{ \sqrt{x} } } \, dx$

Chamando $(1- \sqrt{x} )$ de $u$ e derivando fica:

$du=- \frac{1}{2 \sqrt{x}}dx=>(-2 \sqrt{x}) du=dx$

$\frac{1}{2} \int\limits{ \frac{u^2}{ \sqrt{x} } (-2 \sqrt{x} )du} \, dx$ simplificando as raizes, fica:

$\frac{1}{2} \int\limits{u^2(-2)} \, du=> -\frac{2}{2} \int\limits {u^2} \, du=>-1( \frac{u^3}{3} )=>- \frac{(1- \sqrt{x} )^3}{3}+C$

c) $\int\limits{(x^2+2x- \sqrt{x} )} \, dx => \frac{x^3}{3}+ \frac{2x^2}{2}- \frac{x^{ \frac{3}{2} }}{ \frac{3}{2} } => \frac{x^3}{3}+x^2- \frac{2x^{ \frac{3}{2} }}{3} +C$

Ufa!