Question

Calcule as Integrais Indefinidas

Answer 1

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1)

$\tt{a)}~\Large\boxed{\displaystyle\sf{\int\dfrac{1}{x^2}~dx=-\dfrac{1}{x}+k}}\\\tt{b)}~\boxed{\displaystyle\sf{\int3~dx=3x+k}}\\\tt{c)}~\boxed{\displaystyle\sf{\int(x^5-4)~dx=\dfrac{1}{6}x^6-4x+k}}\\\tt{d)}~\boxed{\displaystyle\sf{\int(x^3-\dfrac{1}{2})~dx}=\dfrac{1}{4}x^4-\dfrac{1}{2}x+k}$

$\tt{e)}~\Large\boxed{\displaystyle\sf{\int(x+1)~dx=\dfrac{1}{2}x^2+x+k}}\\\tt{f)}~\boxed{\displaystyle\sf{\int~dx=x+k}}\\\tt{g)}~\Large\boxed{\displaystyle\sf{\int\left(2+\dfrac{1}{x}\right)~dx=2x+\huge\ell n\left|x\right|+k}}\\\tt{h)}~\boxed{\displaystyle\sf{\int\left(\dfrac{x+3x^2}{x}\right)~dx=\int(1+3x)~dx=x+\dfrac{3}{2}x^2+k}}$

$\tt{i)}~\Large\boxed{\displaystyle\sf{\int e^x~dx=e^x+k}}\\\tt{j)}~\boxed{\displaystyle\sf{\int\sqrt[5]{x^3}~dx=\dfrac{5}{8}x^{\frac{8}{5}}+k}}\\\tt{k)}~\boxed{\displaystyle\sf{\int(x+3e^x)~dx=\dfrac{1}{2}x^2+3e^x+k}}\\\tt{\ell)}~\Large\boxed{\displaystyle\sf{\int cos(x)~dx=sen(x)+k}}$

$\tt{m)}~\Large\boxed{\displaystyle\sf{\int sen(x)~dx=-cos(x)+k}}\\\tt{n)}~\boxed{\displaystyle\sf{\int\left(\dfrac{x^5+3x^2}{x^2}\right)~dx}=\int (x^3+3)~dx=\dfrac{1}{4}x^4+3x+k}\\\tt{o)}~\boxed{\displaystyle\sf{\int\right((2+sen(x)\left)~dx=2x-cos(x)+k}}\\\tt{p)}~\boxed{\displaystyle\sf{\int(x+1)\cdot\sqrt{x}~dx=\int (x^{\frac{3}{2}}+x^{\frac{1}{2}})=\dfrac{2}{5}x^{\frac{5}{2}}+\dfrac{2}{3}x^{\frac{3}{2}+k}}}}}$

$\tt{q)}~\displaystyle\sf{\int\left(x+x^{\frac{1}{3}}\right)\cdot x^2~dx=\int x^3+x^{\frac{7}{3}}~dx}\\\Large\boxed{\displaystyle\sf{\int(x^3+x^{\frac{7}{3}})~dx=\dfrac{1}{4}x^4+\dfrac{3}{10}x^{\frac{10}{3}}}+k}\\\tt{r)}~\huge\boxed{\boxed{\boxed{\boxed{\displaystyle\sf{\int\dfrac{1}{x}~dx=\ell n|x|+k}}}}}$

2)

$\tt{a)}~\sf{f(x)=\sqrt{x}+2x}\\\huge\boxed{\boxed{\boxed{\boxed{\sf{f'(x)=\dfrac{1}{2\sqrt{x}}+2}}}}}\\\tt{b)}~\sf{f(x)=(x^5-2x)^3}\\\sf{f'(x)=3\cdot(x^5-2x)^2\cdot(5x^4-2)}\\\large\boxed{\boxed{\boxed{\boxed{\sf{f'(x)=(15x^4-6)\cdot(x^5-2x)^2}}}}}$