Question

Resolva as integrais:



$\mathsf{A)\int~5a^2x^5~dx }$


$\mathsf{B)~\int~\Big(\sqrt{x}+1\Big)\Big(x-\sqrt{x}+1\Big)dx }$


Por favor , respostas organizadas e detalhadas.
NB: usar códigos LáTex.​

Answer 1

Resposta:

Explicação passo-a-passo:

$a)\int\limits {5a^2x^5}\,dx=5a^2\int\limits {x^5}\,dx =5a^2*\frac{x^6}{6} +c=\frac{5a^2}{6}*x^6+c\\\\b)\int\limits {(\sqrt{x}+1)(x-\sqrt{x}+1)}\,dx=\int\limits{(x\sqrt{x}-x}+\sqrt{x}+x-\sqrt{x}+1)\,dx=\\\\\int\limits{(x.x\frac{1}{2}+1)}\,dx=\int\limits{(x\frac{3}{2}+1)}\,dx=\frac{x\frac{5}{2}}{\frac{5}{2}}+c =\frac{2}{5}*x^{\frac{5}{2}}+c=\frac{2}{5}*\sqrt[]{x^5}+c=\frac{2}{5}*x^2\sqrt{x}+c$

Answer 2

a)

$\int5 {a}^{2} {x}^{5}dx = 5 {a}^{2} \int {x}^{5}dx \\ = 5{a}^{2} . \frac{ {x}^{5 + 1} }{5 + 1} + k \\ = 5 {a}^{2}. \frac{1}{6} {x}^{6} + k$

b)

$\int( \sqrt{x} + 1)(x - \sqrt{x} + 1)dx \\ = \int \: x \sqrt{x} - \int \: xdx + \int \sqrt{x}dx \\ + \int \: xdx - \int \: \sqrt{x}dx + \int \: dx$

$\int( \sqrt{x} + 1)(x - \sqrt{x} + 1)dx \\ = \int \: x \sqrt{x}dx + \int \: dx$

$\int \: x \sqrt{x}dx = \int {x}^{ \frac{3}{2}}dx = \frac{2}{5} {x}^{ \frac{5}{2} } + k$

$\int( \sqrt{x} + 1)(x - \sqrt{x} +1)dx \\ = \frac{2}{5} {x}^{ \frac{5}{2} } + x + k$