Wiskunde Multiple Choice Vraag uit Gricha's Wiskundige Vragenbank

De onbepaalde integraal $\int \frac{1}{(3-x)^5}\:\: dx$ is gelijk aan	A. $\boldsymbol{-\,\frac {1} {4\,(3\,-\,x)^4}+C }$
	B. $\boldsymbol{\frac {1} {6\,(3\,-\,x)^6}+C }$
	C. $\boldsymbol{-\,\ln\|3-x\| + C }$
	D. $\boldsymbol{\frac {1} {4\,(x\,-\,3)^4}+C }$
	E. $\boldsymbol{\frac {1} {5\,(3\,-\,x)^4}+C }$

[ 6-4230 - op net sinds 24.3.14-(E)-15.6.2024 ]

Translation in E N G L I S H

$\int \frac{1}{(3-x)^5}\:\: dx=$ See above

Oplossing - Solution

$https://latex.codecogs.com/png.image?\dpi{110}\int \frac{1}{(3-x)^5}\:\: dx=-\int (3-x)^{-5}d(3-x)=-\frac{(3-x)^{-5+1}}{-5+1}+C\\{\color{Green} .\qquad\qquad\qquad\qquad}=\frac14.\frac{1}{(3-x)^4}+C=\frac{1}{4(x-3)^4}+C$

De onbepaalde integraal $\int \frac{1}{(3-x)^5}\:\: dx$ is gelijk aan	A. \(\boldsymbol{-\,\frac {1} {4\,(3\,-\,x)^4}+C }\)
	B. \(\boldsymbol{\frac {1} {6\,(3\,-\,x)^6}+C }\)
	C. \(\boldsymbol{-\,\ln\|3-x\| + C }\)
	D. \(\boldsymbol{\frac {1} {4\,(x\,-\,3)^4}+C }\)
	E. \(\boldsymbol{\frac {1} {5\,(3\,-\,x)^4}+C }\)