Determine o valor das seguintes integrais.
\int_{\frac{3}{4}}^{\frac{4}{3}} \frac{1}{x \sqrt{1+x^{2}}} d x

Passo 1

O enunciado pede o cálculo da integral
\int_{\frac{3}{4}}^{\frac{4}{3}}\frac{1}{x\sqrt{1+x^2}}\,dx.
O tipo de integral sugere usar uma substituição trigonométrica, pois temos \sqrt{1+x^2}. Vamos usar
x=\tan\theta,\quad dx=\operatorname{sec}^2\theta\,d\theta,
com \theta no intervalo correspondente a x\ge0.

Passo 2

Fazendo x=\tan\theta temos \sqrt{1+x^2}=\operatorname{sec}\theta (com \operatorname{sec}\theta>0) e
\frac{1}{x\sqrt{1+x^2}}\,dx =\frac{1}{\tan\theta\;\operatorname{sec}\theta}\;\operatorname{sec}^2\theta\,d\theta =\frac{\operatorname{sec}^2\theta}{\tan\theta\,\operatorname{sec}\theta}\,d\theta =\frac{\operatorname{sec}\theta}{\tan\theta}\,d\theta =\frac{1/\cos\theta}{\sin\theta/\cos\theta}\,d\theta =\csc\theta\,d\theta.
Quando x=\tfrac34, \theta_1=\arctan\!\bigl(\tfrac34\bigr).
Quando x=\tfrac43, \theta_2=\arctan\!\bigl(\tfrac43\bigr).
Logo a integral original vira
\int_{\theta_1}^{\theta_2}\csc\theta\,d\theta.

Passo 3

A primitiva de \csc\theta é
\int \csc\theta\,d\theta =\ln\bigl|\tan(\tfrac\theta2)\bigr|+C.
Logo,
\int_{\theta_1}^{\theta_2}\csc\theta\,d\theta =\Bigl[\ln\bigl|\tan(\tfrac\theta2)\bigr|\Bigr]_{\theta_1}^{\theta_2} =\ln\bigl|\tan(\tfrac{\theta_2}2)\bigr| -\ln\bigl|\tan(\tfrac{\theta_1}2)\bigr|.

Passo 4

Calculamos as metades dos ângulos usando fórmulas de meia-parte:
Para \theta_1=\arctan\!\bigl(\tfrac34\bigr), temos \sin\theta_1=\tfrac35 e \cos\theta_1=\tfrac45, então
\tan\!\bigl(\tfrac{\theta_1}2\bigr) =\frac{\sin\theta_1}{1+\cos\theta_1} =\frac{\tfrac35}{1+\tfrac45} =\frac{\tfrac35}{\tfrac95} =\tfrac13.
Para \theta_2=\arctan\!\bigl(\tfrac43\bigr), temos \sin\theta_2=\tfrac45 e \cos\theta_2=\tfrac35, então
\tan\!\bigl(\tfrac{\theta_2}2\bigr) =\frac{\sin\theta_2}{1+\cos\theta_2} =\frac{\tfrac45}{1+\tfrac35} =\frac{\tfrac45}{\tfrac85} =\tfrac12.
Substituindo,
\ln\bigl|\tan(\tfrac{\theta_2}2)\bigr| -\ln\bigl|\tan(\tfrac{\theta_1}2)\bigr| =\ln\bigl(\tfrac12\bigr)-\ln\bigl(\tfrac13\bigr) =\ln\!\bigl(\tfrac12\cdot\tfrac{3}{1}\bigr) =\ln\!\bigl(\tfrac32\bigr).

Resposta

\int_{\frac{3}{4}}^{\frac{4}{3}}\frac{1}{x\sqrt{1+x^2}}\,dx =\ln\!\bigl(\tfrac{3}{2}\bigr).

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