Integral de $$$t \sqrt{9 t^{2} + 4}$$$

Encontre $$$\int t \sqrt{9 t^{2} + 4}\, dt$$$.

Seja $$$u=9 t^{2} + 4$$$.

Então $$$du=\left(9 t^{2} + 4\right)^{\prime }dt = 18 t dt$$$ (veja os passos »), e obtemos $$$t dt = \frac{du}{18}$$$.

Logo,

$${\color{red}{\int{t \sqrt{9 t^{2} + 4} d t}}} = {\color{red}{\int{\frac{\sqrt{u}}{18} d u}}}$$

Aplique a regra do múltiplo constante $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ usando $$$c=\frac{1}{18}$$$ e $$$f{\left(u \right)} = \sqrt{u}$$$:

$${\color{red}{\int{\frac{\sqrt{u}}{18} d u}}} = {\color{red}{\left(\frac{\int{\sqrt{u} d u}}{18}\right)}}$$

Aplique a regra da potência $$$\int u^{n}\, du = \frac{u^{n + 1}}{n + 1}$$$ $$$\left(n \neq -1 \right)$$$ com $$$n=\frac{1}{2}$$$:

$$\frac{{\color{red}{\int{\sqrt{u} d u}}}}{18}=\frac{{\color{red}{\int{u^{\frac{1}{2}} d u}}}}{18}=\frac{{\color{red}{\frac{u^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}}}}{18}=\frac{{\color{red}{\left(\frac{2 u^{\frac{3}{2}}}{3}\right)}}}{18}$$

Recorde que $$$u=9 t^{2} + 4$$$:

$$\frac{{\color{red}{u}}^{\frac{3}{2}}}{27} = \frac{{\color{red}{\left(9 t^{2} + 4\right)}}^{\frac{3}{2}}}{27}$$

Portanto,

$$\int{t \sqrt{9 t^{2} + 4} d t} = \frac{\left(9 t^{2} + 4\right)^{\frac{3}{2}}}{27}$$

Adicione a constante de integração:

$$\int{t \sqrt{9 t^{2} + 4} d t} = \frac{\left(9 t^{2} + 4\right)^{\frac{3}{2}}}{27}+C$$

$$$\int t \sqrt{9 t^{2} + 4}\, dt = \frac{\left(9 t^{2} + 4\right)^{\frac{3}{2}}}{27} + C$$$A