Resolver las siguientes integrales indefinidas ...

Enunciado:
Resolver las siguientes integrales indefinidas:

  a) $\displaystyle \int x \,\ln x \,dx$

  b) $\displaystyle \int \sqrt{2x+1} \, dx$

  c) $\displaystyle \int \dfrac{2x}{x^2+1} \,dx$

  d) $\displaystyle \int 2^x \,dx$

Resolución:

a)
Integrando por partes ( $\int u\,dv=u\,v-\int v\,du$ ) y haciendo las siguientes asignaciones:
$u:=\ln x \Rightarrow du = \dfrac{1}{x}\,dx$ y $dv:=x\,dx \Rightarrow v= \dfrac{1}{2}\,x^2$
podemos escribir la integral pedida de la forma
$\displaystyle \int x \,\ln x \,dx = \dfrac{1}{2}\,x^2\,\ln x - \int \, \dfrac{1}{2}\,x^2 \, \dfrac{1}{x}\,dx $
e integrando - fácilmente ya - la integral del segundo término llegamos a
$\displaystyle \int x \,\ln x \,dx = \dfrac{1}{2}\,x^2\,\ln x - \dfrac{1}{4}\,x^2 + C = \dfrac{1}{2}\,x^2 \, ( \ln x - \dfrac{1}{2} ) + C $

b)
$\displaystyle \int \sqrt{2x+1} \, dx \underset{(1)}{=} \int \sqrt{2x+1} \, d\big( \frac{1}{2}\,(2x+1) \big)=\dfrac{1}{2}\,\int (2x+1)^{\frac{1}{2}} \, d(2x+1)$
    $=\displaystyle \dfrac{1}{2}\,\dfrac{1}{\frac{1}{2}+1}\,(2x+1)^{\frac{1}{2}+1}+C$
    $=\displaystyle \frac{1}{2}\cdot \frac{2}{3}\,(2x+1)^{\frac{3}{2}}+C$
    $=\displaystyle \dfrac{1}{3}\,(2x+1)\,\sqrt{2x+1}+C$

---
(1)     $d\,\big(\dfrac{1}{2}\,(2\,x+1)\big)=\big(\dfrac{1}{2}\,(2\,x+1)\big)'\,dx=\dfrac{1}{2}\cdot 2\,dx=dx$
---

c)
$\displaystyle \int \dfrac{2x}{x^2+1} \,dx = \int \dfrac{2x\,dx}{x^2+1} \underset{(2)}{=} \int \dfrac{d(x^2+1)}{x^2+1}=\ln ( x^2+1)+C $

(2)     $d(x^2+1)=(x^2+1)'\,dx=2x\,dx$

d)
$\displaystyle \int 2^x \,dx = \int \, e^{x\,\ln 2} \,dx \underset{(3)}{=} \int \, d\big( \dfrac{1}{\ln 2}\,e^{x\,\ln 2} \big) = \dfrac{1}{\ln 2}\,e^{x\,\ln 2} + C = \dfrac{1}{\ln 2}\,2^{x} +C$

(3)   $d\big( \dfrac{1}{\ln 2}\,e^{x\,\ln 2} \big) = \big( \dfrac{1}{\ln 2}\,e^{x\,\ln 2} \big)'\,dx = e^{x\,\ln 2} \,dx$

$\square$

[nota del autor]