Indefinite integral practice
While watching the video, I came across an interesting question, so I would like to share it with you here.
topic
Calculate ∫ ( 1 + x − 1 x ) ex + 1 xdx \int(1+x-\dfrac 1x)e^{x+\frac 1x}dx∫(1+x−x1)ex+x1dx
Solution:
\qquad原式 = ∫ e x + 1 x d x + ∫ x ( 1 − 1 x 2 ) e x + 1 x d x =\int e^{x+\frac 1x}dx+\int x(1-\dfrac{1}{x^2})e^{x+\frac 1x}dx =∫ex+x1dx+∫x(1−x21)ex+x1dx
= ∫ e x + 1 x d x + ∫ x d ( e x + 1 x ) \qquad\qquad =\int e^{x+\frac 1x}dx+\int xd(e^{x+\frac 1x}) =∫ex+x1dx+∫x d ( ex+x1)
= ∫ e x + 1 x d x + x e x + 1 x − ∫ e x + 1 x d x \qquad\qquad =\int e^{x+\frac 1x}dx+xe^{x+\frac 1x}-\int e^{x+\frac 1x}dx =∫ex+x1dx+x ex+x1−∫ex+x1dx
= x e x + 1 x \qquad\qquad =xe^{x+\frac 1x} =x ex+x1
This question examines the integration by parts method . First divide the formula into two parts, and then divide ∫ x ( 1 − 1 x 2 ) ex + 1 xdx \int x(1-\dfrac{1}{x^2}) e^{x+\frac 1x}dx∫x(1−x21)ex+x1dx看作 u ( x ) = x , v ( x ) = e x + 1 x u(x)=x,v(x)=e^{x+\frac 1x} u(x)=x,v(x)=ex+x1Integrate by parts, and finally offset the two parts to get the answer.