Calcule a integral usando integração por partes.
  • \int\left(x^{2}+1\right) e^{-x} d x
  • \int e^{-\theta} \cos (2 \theta) d \theta
  • \int \frac{\ln x}{x^{3}} d x
  • \int \arctan \left(\frac{1}{x}\right) d x

Passo 1

O enunciado pede para calcular quatro integrais usando integração por partes, cuja fórmula é
\int u\,dv = u\,v \;-\;\int v\,du.
As integrais são
\displaystyle \int (x^{2}+1)e^{-x}\,dx, \displaystyle \int e^{-\theta}\cos(2\theta)\,d\theta, \displaystyle \int \frac{\ln x}{x^{3}}\,dx
e
\displaystyle \int \arctan\!\bigl(\tfrac1x\bigr)\,dx.

Passo 2

Para \displaystyle \int (x^{2}+1)e^{-x}\,dx separemos
\int (x^{2}+1)e^{-x}\,dx =\int x^{2}e^{-x}\,dx \;+\;\int e^{-x}\,dx.
1. Em \displaystyle \int x^{2}e^{-x}\,dx tomamos u=x^{2}, dv=e^{-x}dx.
Então du=2x\,dx, v=-e^{-x}, e
\int x^{2}e^{-x}\,dx = -x^{2}e^{-x} - \int(-e^{-x})\cdot2x\,dx = -x^{2}e^{-x} +2\int x e^{-x}\,dx.
2. Em \displaystyle \int x e^{-x}\,dx tomamos u=x, dv=e^{-x}dx,
daí du=dx, v=-e^{-x}, e
\int x e^{-x}\,dx = -x e^{-x} - \int(-e^{-x})\,dx = -x e^{-x} - e^{-x}.
Logo
\displaystyle \int x^{2}e^{-x}\,dx = -x^{2}e^{-x} +2\bigl(-x e^{-x}-e^{-x}\bigr) = -(x^{2}+2x+2)e^{-x}. 3. Por fim
\int (x^{2}+1)e^{-x}\,dx = -(x^{2}+2x+2)e^{-x} + \int e^{-x}\,dx = -(x^{2}+2x+2)e^{-x} - e^{-x} = -(x^{2}+2x+3)e^{-x}.

Passo 3

Para \displaystyle \int e^{-\theta}\cos(2\theta)\,d\theta definamos
I=\int e^{-\theta}\cos(2\theta)\,d\theta,\quad J=\int e^{-\theta}\sin(2\theta)\,d\theta.
1. Com u=\cos(2\theta), dv=e^{-\theta}d\theta vem
\,du=-2\sin(2\theta)d\theta,\;v=-e^{-\theta}, então
I = -e^{-\theta}\cos(2\theta) - \int(-e^{-\theta})(-2\sin(2\theta))\,d\theta = -e^{-\theta}\cos(2\theta) - 2J.
2. Para J tomamos u=\sin(2\theta), dv=e^{-\theta}d\theta,
obtendo
J = -e^{-\theta}\sin(2\theta) + 2I.
3. Substituindo J em I:
I = -e^{-\theta}\cos(2\theta) -2\bigl(-e^{-\theta}\sin(2\theta)+2I\bigr) = -e^{-\theta}\cos(2\theta) +2e^{-\theta}\sin(2\theta) -4I.
Logo 5I = e^{-\theta}\bigl(2\sin(2\theta)-\cos(2\theta)\bigr) e
I = \frac{e^{-\theta}}{5}\bigl(2\sin(2\theta)-\cos(2\theta)\bigr).

Passo 4

Para \displaystyle \int \frac{\ln x}{x^{3}}\,dx escrevemos x^{-3}\ln x.
Tomamos u=\ln x, dv=x^{-3}dx. Então
\,du=\frac{dx}{x},\;v=\int x^{-3}dx = -\frac1{2x^{2}}, e
\int x^{-3}\ln x\,dx = u\,v - \int v\,du = -\frac{\ln x}{2x^{2}} - \int\Bigl(-\frac1{2x^{2}}\Bigr)\frac{dx}{x} = -\frac{\ln x}{2x^{2}} + \frac12\int x^{-3}dx.
Como \displaystyle \int x^{-3}dx = -\frac1{2x^{2}}, obtemos
\int \frac{\ln x}{x^{3}}\,dx = -\frac{\ln x}{2x^{2}} - \frac1{4x^{2}} = -\frac{2\ln x +1}{4x^{2}}.

Passo 5

Para \displaystyle \int \arctan\!\bigl(\tfrac1x\bigr)\,dx usemos
u=\arctan\bigl(\tfrac1x\bigr), dv=dx. Então
\,v=x e
du = \frac{1}{1+(1/x)^{2}}\cdot\bigl(-\tfrac1{x^{2}}\bigr)\,dx = -\frac{dx}{x^{2}+1}.
Aplicando
\int u\,dv = x\arctan\!\bigl(\tfrac1x\bigr) - \int x\Bigl(-\frac{1}{x^{2}+1}\Bigr)\,dx = x\arctan\!\bigl(\tfrac1x\bigr) + \int \frac{x}{x^{2}+1}\,dx.
Com t=x^{2}+1, dt=2x\,dx, vem
\displaystyle \int \frac{x}{x^{2}+1}\,dx = \tfrac12\ln\!\lvert x^{2}+1\rvert.
Logo
\int \arctan\!\bigl(\tfrac1x\bigr)\,dx = x\arctan\!\bigl(\tfrac1x\bigr) + \frac12\ln\bigl(x^{2}+1\bigr).

Resposta

\int (x^{2}+1)e^{-x}\,dx = -\bigl(x^{2}+2x+3\bigr)e^{-x}+C.
\int e^{-\theta}\cos(2\theta)\,d\theta = \frac{e^{-\theta}}{5}\bigl(2\sin(2\theta)-\cos(2\theta)\bigr)+C.
\int \frac{\ln x}{x^{3}}\,dx = -\frac{2\ln x +1}{4x^{2}}+C.
\int \arctan\!\bigl(\tfrac1x\bigr)\,dx = x\arctan\!\bigl(\tfrac1x\bigr) + \tfrac12\ln\bigl(x^{2}+1\bigr)+C.

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