Calcule as integrais por substituição trigonométrica:
\int \sqrt{4+x^{2}} d x
\frac{\sqrt{4+x^{2}}}{\alpha} x \quad
\operatorname{tg} \alpha=x / 2
\alpha=\operatorname{arctg} x / 2
x=2 \cdot \operatorname{tg} \alpha
\frac{d x}{d \alpha}=2 \cdot \operatorname{sec} ^{2} \alpha \Rightarrow d x=2 \operatorname{sec} ^{2} \alpha d \alpha
\operatorname{sec} \alpha= & \frac{\sqrt{4+x^{2}}}{2}

Passo 1

O enunciado pede para calcular
\int \sqrt{4+x^{2}}\;dx
usando substituição trigonométrica. Vamos usar a substituição
x = 2\tan\alpha,\quad dx = 2\operatorname{sec}^{2}\!\alpha\,d\alpha,
e observar que
\sqrt{4+x^{2}} = \sqrt{4+4\tan^{2}\!\alpha} = 2\operatorname{sec}\alpha.

Passo 2

Substituindo na integral original, temos
\int \sqrt{4+x^{2}}\;dx = \int \bigl(2\operatorname{sec}\alpha\bigr)\,\bigl(2\operatorname{sec}^{2}\!\alpha\,d\alpha\bigr) = 4\int \operatorname{sec}^{3}\!\alpha\,d\alpha.

Passo 3

Para calcular \displaystyle \int \operatorname{sec}^{3}\!\alpha\,d\alpha, usamos integração por partes. Escrevemos
\int \operatorname{sec}^{3}\!\alpha\,d\alpha = \int \operatorname{sec}\alpha\cdot\operatorname{sec}^{2}\!\alpha\,d\alpha
e tomamos
u = \operatorname{sec}\alpha,\quad dv = \operatorname{sec}^{2}\!\alpha\,d\alpha \implies du = \operatorname{sec}\alpha\tan\alpha\,d\alpha,\quad v = \tan\alpha.
Então
\int \operatorname{sec}^{3}\!\alpha\,d\alpha = \operatorname{sec}\alpha\tan\alpha - \int \tan\alpha\cdot\operatorname{sec}\alpha\tan\alpha\,d\alpha = \operatorname{sec}\alpha\tan\alpha - \int \operatorname{sec}\alpha\tan^{2}\!\alpha\,d\alpha.
Como \tan^{2}\!\alpha = \operatorname{sec}^{2}\!\alpha - 1, segue
\int \operatorname{sec}^{3}\!\alpha\,d\alpha = \operatorname{sec}\alpha\tan\alpha - \int \bigl(\operatorname{sec}^{3}\!\alpha - \operatorname{sec}\alpha\bigr)\,d\alpha = \operatorname{sec}\alpha\tan\alpha - \int \operatorname{sec}^{3}\!\alpha\,d\alpha + \int \operatorname{sec}\alpha\,d\alpha.
Reorganizando,
2\int \operatorname{sec}^{3}\!\alpha\,d\alpha = \operatorname{sec}\alpha\tan\alpha + \int \operatorname{sec}\alpha\,d\alpha = \operatorname{sec}\alpha\tan\alpha + \ln\bigl|\operatorname{sec}\alpha + \tan\alpha\bigr|.
Logo
\int \operatorname{sec}^{3}\!\alpha\,d\alpha = \tfrac12\bigl(\operatorname{sec}\alpha\tan\alpha + \ln\bigl|\operatorname{sec}\alpha + \tan\alpha\bigr|\bigr).

Passo 4

Voltando à integral multiplicada por 4, obtemos
4\int \operatorname{sec}^{3}\!\alpha\,d\alpha = 4\cdot \tfrac12\bigl(\operatorname{sec}\alpha\tan\alpha + \ln\bigl|\operatorname{sec}\alpha + \tan\alpha\bigr|\bigr) = 2\operatorname{sec}\alpha\tan\alpha + 2\ln\bigl|\operatorname{sec}\alpha + \tan\alpha\bigr| + C.
Agora substituímos \operatorname{sec}\alpha = \dfrac{\sqrt{4+x^{2}}}{2} e \tan\alpha = \dfrac{x}{2}:
2\operatorname{sec}\alpha\tan\alpha = 2\cdot\frac{\sqrt{4+x^{2}}}{2}\cdot\frac{x}{2} = \frac{x\sqrt{4+x^{2}}}{2},
\operatorname{sec}\alpha + \tan\alpha = \frac{\sqrt{4+x^{2}}}{2} + \frac{x}{2} = \frac{x + \sqrt{4+x^{2}}}{2}.
Como \ln\bigl|\tfrac{x+\sqrt{4+x^{2}}}{2}\bigr| = \ln\bigl|x+\sqrt{4+x^{2}}\bigr| - \ln 2, o termo -2\ln 2 entra em C.

Resposta

\int \sqrt{4+x^{2}}\;dx = \frac{x\sqrt{4+x^{2}}}{2} + 2\ln\bigl|x + \sqrt{4+x^{2}}\bigr| + C.

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