Calcule as integrais dos exercícios 35-40.
40.
\int \frac{(x+1)^{2} \operatorname{arctg}(3 x)+9 x^{3}+x}{\left(9 x^{2}+1\right)(x+1)^{2}} d x

Passo 1

O enunciado pede calcular
\int \frac{(x+1)^{2}\,\operatorname{arctg}(3x)+9x^3+x}{\bigl(9x^2+1\bigr)\,(x+1)^2}\,dx.
Vamos reconhecer a derivada de \operatorname{arctg}(3x) e também simplificar a parte racional por decomposição.

Passo 2

Separemos o integrando em duas parcelas:
\frac{(x+1)^2\,\operatorname{arctg}(3x)}{\bigl(9x^2+1\bigr)(x+1)^2} +\frac{9x^3+x}{\bigl(9x^2+1\bigr)(x+1)^2} =\frac{\operatorname{arctg}(3x)}{9x^2+1}+\frac{x\,(9x^2+1)}{\bigl(9x^2+1\bigr)(x+1)^2} =\frac{\operatorname{arctg}(3x)}{9x^2+1}+\frac{x}{(x+1)^2}.
Logo
I=\int\frac{\operatorname{arctg}(3x)}{9x^2+1}\,dx \;+\;\int\frac{x}{(x+1)^2}\,dx.

Passo 3

Para
I_1=\int\frac{\operatorname{arctg}(3x)}{9x^2+1}\,dx
façamos u=\operatorname{arctg}(3x). Então
du=\frac{3}{9x^2+1}\,dx \quad\Longrightarrow\quad \frac{dx}{9x^2+1}=\frac{du}{3}.
Assim
I_1=\int u\;\frac{du}{3}=\frac{1}{3}\int u\,du =\frac{u^2}{6} =\frac{\bigl(\operatorname{arctg}(3x)\bigr)^2}{6}.

Passo 4

Para
I_2=\int\frac{x}{(x+1)^2}\,dx
escrevemos x=(x+1)-1, logo
\frac{x}{(x+1)^2} =\frac{x+1}{(x+1)^2}-\frac{1}{(x+1)^2} =\frac{1}{x+1}-\frac{1}{(x+1)^2}.
Portanto
I_2 =\int\frac{1}{x+1}\,dx \;-\;\int\frac{1}{(x+1)^2}\,dx =\ln\lvert x+1\rvert +\frac{1}{x+1}.

Passo 5

Somando I_1+I_2 concluímos que
\int \frac{(x+1)^{2}\,\operatorname{arctg}(3x)+9x^3+x}{\bigl(9x^2+1\bigr)\,(x+1)^2}\,dx =\frac{\bigl(\operatorname{arctg}(3x)\bigr)^2}{6} +\ln\lvert x+1\rvert +\frac{1}{x+1} +C.

Resposta

\frac{\bigl(\operatorname{arctg}(3x)\bigr)^2}{6} +\ln\lvert x+1\rvert +\frac{1}{x+1} +C.

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