16.2 Problemas

  1. Repetindo procedimento análogo ao usado no exemplo 16.5 , mostre que
\int \sqrt{x^{2}+\lambda} d x=\frac{x}{2} \sqrt{x^{2}+\lambda}+\frac{\lambda}{2} \ln \left|x+\sqrt{x^{2}+\lambda}\right|+C
  1. Calcule as seguintes integrais.
    (a) \int x e^{x} d x. Resposta. e^{x}(x-1)+C.
    (b) \int \ln x d x. Resposta. x(\ln x-1)+C.
    (c) \int x^{n} \ln x d x(n \neq-1). Resposta. \frac{x^{n+1}}{n+1}\left(\ln x-\frac{1}{n+1}\right)+C.
    (d) \int \ln \left(1+x^{2}\right) \mathrm{dx}. Resposta. x \ln \left(x^{2}+1\right)-2 x+2 \operatorname{arctg} x+\mathrm{C}.
    (e) \int x \operatorname{arctg} x d x. Resposta. \frac{1}{2}\left[\left(x^{2}+1\right) \operatorname{arctg} x-x\right]+C.
    (f) \int \operatorname{arcsen} x d x. Resposta. x \operatorname{arcsen} x+\sqrt{1-x^{2}}+C.
    (g) \int \sqrt{1-x^{2}} \mathrm{dx}. Resposta. \frac{1}{2} \operatorname{arcsen} x+\frac{x}{2} \sqrt{1-x^{2}}+\mathrm{C}.

Passo 1

O enunciado pede duas coisas principais:
  1. Mostrar por partes que
\int \sqrt{x^{2}+\lambda}\,dx=\frac{x}{2}\sqrt{x^{2}+\lambda}+\frac{\lambda}{2}\ln\bigl|x+\sqrt{x^{2}+\lambda}\bigr|+C.
  1. Calcular várias integrais usando técnicas como integração por partes e substituição trigonométrica. Vamos assim recorrer a formulários básicos de partes ( \int u\,dv = u\,v - \int v\,du) e, quando necessário, a substituições adequadas.

Passo 2

Seja
I=\int \sqrt{x^{2}+\lambda}\,dx.
Aplicamos integração por partes escolhendo
u=\sqrt{x^{2}+\lambda},\quad dv=dx.
Então
du=\frac{x}{\sqrt{x^{2}+\lambda}}\,dx,\quad v=x,
e
I = x\sqrt{x^{2}+\lambda} \;-\;\int x\;\frac{x}{\sqrt{x^{2}+\lambda}}\,dx = x\sqrt{x^{2}+\lambda} \;-\;\int \frac{x^{2}}{\sqrt{x^{2}+\lambda}}\,dx.
Escrevendo x^{2}=(x^{2}+\lambda)-\lambda, temos
I = x\sqrt{x^{2}+\lambda} -\int \sqrt{x^{2}+\lambda}\,dx +\lambda\int \frac{dx}{\sqrt{x^{2}+\lambda}} = x\sqrt{x^{2}+\lambda}-I +\lambda\int \frac{dx}{\sqrt{x^{2}+\lambda}}.
Somando I aos dois lados surge
2I = x\sqrt{x^{2}+\lambda} +\lambda\int \frac{dx}{\sqrt{x^{2}+\lambda}}.
Sabemos que
\int \frac{dx}{\sqrt{x^{2}+\lambda}} =\ln\bigl|x+\sqrt{x^{2}+\lambda}\bigr|+C,
logo
I=\frac{x}{2}\sqrt{x^{2}+\lambda} +\frac{\lambda}{2}\ln\bigl|x+\sqrt{x^{2}+\lambda}\bigr|+C.

Passo 3

Calculemos \displaystyle\int x e^{x}dx por partes, escolhendo
u=x,\quad dv=e^{x}dx\quad\Longrightarrow\quad du=dx,\;v=e^{x}.
Então
\int x e^{x}dx = x e^{x}-\int e^{x}dx = x e^{x}-e^{x}+C = e^{x}(x-1)+C.

Passo 4

Para \displaystyle\int \ln x\,dx use partes com
u=\ln x,\quad dv=dx\quad\Longrightarrow\quad du=\frac{dx}{x},\;v=x.
Obtemos
\int \ln x\,dx = x\ln x-\int x\frac{dx}{x} = x\ln x-\int 1\,dx = x\ln x - x +C = x(\ln x-1)+C.

Passo 5

Em \displaystyle\int x^{n}\ln x\,dx (com n\neq -1) escolhemos
u=\ln x,\quad dv=x^{n}dx \quad\Longrightarrow\quad du=\frac{dx}{x},\;v=\frac{x^{n+1}}{n+1}.
Logo
\int x^{n}\ln x\,dx =\frac{x^{n+1}}{n+1}\ln x -\int \frac{x^{n+1}}{n+1}\cdot\frac{dx}{x} =\frac{x^{n+1}}{n+1}\ln x -\frac{1}{n+1}\int x^{n}\,dx =\frac{x^{n+1}}{n+1}\Bigl(\ln x-\frac{1}{n+1}\Bigr)+C.

Passo 6

Para \displaystyle\int \ln(1+x^{2})\,dx use partes com
u=\ln(1+x^{2}),\quad dv=dx \quad\Longrightarrow\quad du=\frac{2x}{1+x^{2}}\,dx,\;v=x.
Temos
\int \ln(1+x^{2})\,dx = x\ln(1+x^{2}) -\int x\cdot\frac{2x}{1+x^{2}}\,dx = x\ln(1+x^{2}) -2\int \frac{x^{2}}{1+x^{2}}\,dx.
Note que \frac{x^{2}}{1+x^{2}}=1-\frac{1}{1+x^{2}}, então
\int \ln(1+x^{2})\,dx = x\ln(1+x^{2}) -2\int\!1\,dx +2\int\frac{dx}{1+x^{2}} = x\ln(1+x^{2})-2x+2\arctan x +C.

Passo 7

Em \displaystyle\int x\arctan x\,dx use partes com
u=\arctan x,\quad dv=x\,dx \quad\Longrightarrow\quad du=\frac{dx}{1+x^{2}},\;v=\frac{x^{2}}{2}.
Assim
\int x\arctan x\,dx =\frac{x^{2}}{2}\arctan x -\int \frac{x^{2}}{2}\cdot\frac{dx}{1+x^{2}} =\frac{x^{2}+1}{2}\arctan x -\frac{1}{2}\int \frac{1+x^{2}}{1+x^{2}}\,dx -\frac{1}{2}\int \frac{-1}{1+x^{2}}\,dx.
Mais diretamente, escrevendo \frac{x^{2}}{1+x^{2}}=1-\frac{1}{1+x^{2}}, resulta
\int x\arctan x\,dx =\frac{(x^{2}+1)\arctan x}{2} -\frac{1}{2}\int 1\,dx +\frac{1}{2}\int\frac{dx}{1+x^{2}} =\frac{(x^{2}+1)\arctan x - x}{2} +\frac{1}{2}\arctan x +C,
mas reunindo fica
\frac{(x^{2}+1)\arctan x - x}{2}+C.

Passo 8

Para \displaystyle\int \arcsin x\,dx use partes com
u=\arcsin x,\quad dv=dx \quad\Longrightarrow\quad du=\frac{dx}{\sqrt{1-x^{2}}},\;v=x.
Logo
\int \arcsin x\,dx = x\arcsin x -\int \frac{x}{\sqrt{1-x^{2}}}\,dx.
Na segunda integral faça a substituição t=1-x^{2}, dt=-2x\,dx, obtendo
\int \frac{x}{\sqrt{1-x^{2}}}\,dx=-\frac{1}{2}\int t^{-1/2}\,dt =-\sqrt{t} =-\sqrt{1-x^{2}}.
Portanto
\int \arcsin x\,dx = x\arcsin x + \sqrt{1-x^{2}} +C.

Passo 9

Para \displaystyle\int \sqrt{1-x^{2}}\,dx use a substituição trigonométrica
x=\sin\theta,\quad dx=\cos\theta\,d\theta,\quad\sqrt{1-x^{2}}=\cos\theta.
Isso dá
\int \sqrt{1-x^{2}}\,dx =\int \cos^{2}\theta\,d\theta =\int \frac{1+\cos(2\theta)}{2}\,d\theta =\frac{\theta}{2}+\frac{\sin(2\theta)}{4}+C.
De volta a x, temos \theta=\arcsin x e \sin(2\theta)=2\sin\theta\cos\theta=2x\sqrt{1-x^{2}}. Logo
\int \sqrt{1-x^{2}}\,dx =\frac{\arcsin x}{2} + \frac{x\sqrt{1-x^{2}}}{2} +C.

Resposta

  1. \displaystyle\int \sqrt{x^{2}+\lambda}\,dx =\frac{x}{2}\sqrt{x^{2}+\lambda} +\frac{\lambda}{2}\ln\bigl|x+\sqrt{x^{2}+\lambda}\bigr|+C.
(a) \displaystyle\int x e^{x}dx = e^{x}(x-1)+C.
(b) \displaystyle\int \ln x\,dx = x(\ln x-1)+C.
(c) \displaystyle\int x^{n}\ln x\,dx =\frac{x^{n+1}}{n+1}\Bigl(\ln x-\frac{1}{n+1}\Bigr)+C,\;n\neq-1.
(d) \displaystyle\int \ln(1+x^{2})\,dx =x\ln(1+x^{2})-2x+2\arctan x+C.
(e) \displaystyle\int x\arctan x\,dx =\frac{(x^{2}+1)\arctan x - x}{2}+C.
(f) \displaystyle\int \arcsin x\,dx =x\arcsin x+\sqrt{1-x^{2}}+C.
(g) \displaystyle\int \sqrt{1-x^{2}}\,dx =\frac{\arcsin x}{2}+\frac{x\sqrt{1-x^{2}}}{2}+C.

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