Use as seguintes fórmulas dadas para calcular as integrais a seguir.
  • \int \frac{1}{a^{2}+x^{2}} d x=\frac{1}{a} \operatorname{arctg} \frac{x}{a}+C.
  • \int \frac{1}{\sqrt{a^{2}-x^{2}}} d x=\operatorname{arcsen} \frac{x}{a}+C,|x|<a.
  • \int \frac{1}{x \sqrt{x^{2}-a^{2}}} d x=\frac{1}{a} \operatorname{arcsec}\left|\frac{x}{a}\right|+C,|x|>a.
    a) \int \frac{1}{1+4 x^{2}} d x.
    b) \int_{0}^{2} \frac{1}{x^{2}+4} d x.
    c) \int \frac{x^{2}}{1+x^{6}} d x.
    d) \int \frac{e^{x}}{\sqrt{1-e^{2 x}}} d x.
    e) \int \frac{\cos x}{\sqrt{4-\operatorname{sen}^{2} x}} d x.
    f) \int \frac{1}{e^{x} \sqrt{1-e^{-2 x}}} d x.
    g) \int \frac{1}{\sqrt{e^{2 x}-1}} d x.
    h) \int_{\frac{4 \sqrt{3}}{3}}^{4} \frac{1}{x \sqrt{x^{2}-4}} d x.

Passo 1

O enunciado pede para calcular várias integrais usando as fórmulas padrão para \displaystyle\int\frac{1}{a^{2}+x^{2}}dx, \displaystyle\int\frac{1}{\sqrt{a^{2}-x^{2}}}dx e \displaystyle\int\frac{1}{x\sqrt{x^{2}-a^{2}}}dx.
Vamos usar substituições simples para reduzir cada caso a uma dessas formas.

Passo 2

a) Para \displaystyle\int\frac{1}{1+4x^{2}}dx escrevemos 1+4x^{2}=1+(2x)^{2}. Fazer u=2x leva a du=2\,dx, ou dx=\tfrac{du}{2}, então
\int\frac{1}{1+4x^{2}}dx =\int\frac{1}{1+u^{2}}\cdot\frac{du}{2} =\frac{1}{2}\arctan(u)+C =\frac{1}{2}\arctan(2x)+C.
b) Para \displaystyle\int_{0}^{2}\frac{1}{x^{2}+4}dx usamos o mesmo primitivo com 4=2^{2}: \displaystyle F(x)=\frac{1}{2}\arctan\bigl(\tfrac{x}{2}\bigr).
Logo
\int_{0}^{2}\frac{1}{x^{2}+4}dx =F(2)-F(0) =\frac{1}{2}\bigl[\arctan(1)-\arctan(0)\bigr] =\frac{1}{2}\cdot\frac{\pi}{4} =\frac{\pi}{8}.

Passo 3

c) Em \displaystyle\int\frac{x^{2}}{1+x^{6}}dx, note que 1+x^{6}=1+(x^{3})^{2}. Fazendo u=x^{3} du=3x^{2}dx, ou x^{2}dx=\tfrac{du}{3}. Então
\int\frac{x^{2}}{1+x^{6}}dx =\frac{1}{3}\int\frac{1}{1+u^{2}}du =\frac{1}{3}\arctan(u)+C =\frac{1}{3}\arctan(x^{3})+C.

Passo 4

d) Para \displaystyle\int\frac{e^{x}}{\sqrt{1-e^{2x}}}dx, faça u=e^{x}, du=e^{x}dx. A integral fica
\int\frac{du}{\sqrt{1-u^{2}}} =\arcsin(u)+C =\arcsin\bigl(e^{x}\bigr)+C.

Passo 5

e) Em \displaystyle\int\frac{\cos x}{\sqrt{4-\sin^{2}x}}dx, use u=\sin x, du=\cos x\,dx. Como 4-\sin^{2}x=4-u^{2},
\int\frac{du}{\sqrt{4-u^{2}}} =\arcsin\!\Bigl(\frac{u}{2}\Bigr)+C =\arcsin\!\Bigl(\frac{\sin x}{2}\Bigr)+C.

Passo 6

f) Para \displaystyle\int\frac{1}{e^{x}\sqrt{1-e^{-2x}}}dx escreva \frac{1}{e^{x}}dx=e^{-x}dx e note que \sqrt{1-e^{-2x}} fica com u=e^{-x}, du=-e^{-x}dx. Assim
\int\frac{e^{-x}dx}{\sqrt{1-e^{-2x}}} =-\int\frac{du}{\sqrt{1-u^{2}}} =-\arcsin(u)+C =-\arcsin\bigl(e^{-x}\bigr)+C.

Passo 7

g) Em \displaystyle\int\frac{1}{\sqrt{e^{2x}-1}}dx, use u=e^{x}, du=e^{x}dx, então dx=\tfrac{du}{u} e
\int\frac{1}{\sqrt{u^{2}-1}}\cdot\frac{du}{u} =\int\frac{du}{u\sqrt{u^{2}-1}} =\arcsec\bigl|u\bigr|+C =\arcsec\bigl(e^{x}\bigr)+C.

Passo 8

h) Para \displaystyle\int_{\frac{4\sqrt{3}}{3}}^{4}\frac{1}{x\sqrt{x^{2}-4}}dx usamos a fórmula \displaystyle\int\frac{dx}{x\sqrt{x^{2}-a^{2}}}=\frac{1}{a}\arcsec\!\bigl|\tfrac{x}{a}\bigr|+C com a=2. O primitivo é \displaystyle \frac{1}{2}\arcsec\!\bigl(\tfrac{x}{2}\bigr). Avaliando:
\frac{1}{2}\Bigl[\arcsec\!(4/2)-\arcsec\!\bigl(\tfrac{4\sqrt{3}/3}{2}\bigr)\Bigr] =\frac{1}{2}\Bigl[\arcsec(2)-\arcsec\bigl(\tfrac{2\sqrt{3}}{3}\bigr)\Bigr] =\frac{1}{2}\Bigl[\tfrac{\pi}{3}-\tfrac{\pi}{6}\Bigr] =\frac{\pi}{12}.

Resposta

a) \displaystyle\frac{1}{2}\arctan(2x)+C
b) \displaystyle\frac{\pi}{8}
c) \displaystyle\frac{1}{3}\arctan(x^{3})+C
d) \displaystyle\arcsin\bigl(e^{x}\bigr)+C
e) \displaystyle\arcsin\!\bigl(\tfrac{\sin x}{2}\bigr)+C
f) \displaystyle -\arcsin\bigl(e^{-x}\bigr)+C
g) \displaystyle\arcsec\bigl(e^{x}\bigr)+C
h) \displaystyle\frac{\pi}{12}

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