Papula.integreren.substitutie Tabel

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INTEGRATIE MET SUBSTITUTIE type integraal substitutie oplossing voorbeelden A. dx b x a f u = a x + b a dx du a du dx 1. dx 3 x 2 3 2. dx 5 x 4 3. dx e 2 x 3 1. u = 2 x – 3 2. u = 4 x + 5 3. u = 3 x + 2 B. dx x ' f . x f u = f (x) x ' f dx du f'(x).dx = du ½ f 2 (x) + C 1. dx x cos . x sin 2. dx x x ln 1. u = sin x 2. u = ln x C. dx x f x ' f u = f (x) x ' f dx du f'(x).dx = du ln |f (x)| + C 1. dx 1 x 3 x 3 x 2 2 2. dx 5 e e x x 3. dx x cos x sin 1. u = x 2 – 3 x + 1 2. u = e x + 5 3. u = cos x D. dx x a , x f 2 2 x = a.Sin u u cos . a du dx dx = a.cos u.du let op : (a 2 – x 2) =a.cosu 1. dx x r 2 2 2. dx x r . x 2 2 3. 2 x 4 dx 1. x = r.sin u 2. x = r.sin u 3. x = 2.sin u

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Papula.integreren.substitutie Tabel

Transcript of Papula.integreren.substitutie Tabel

Page 1: Papula.integreren.substitutie Tabel

INTEGRATIE MET SUBSTITUTIE type integraal substitutie oplossing voorbeelden

A. dxbxaf u = a x + b

adxdu

adu

dx

1. dx3x2 3

2. dx5x4

3. dxe 2x3

1. u = 2 x – 3 2. u = 4 x + 5 3. u = 3 x + 2

B. dxx'f.xf u = f (x)

x'fdxdu

f'(x).dx = du

½ f2 (x) + C 1. dxxcos.xsin

2. dxxxln

1. u = sin x 2. u = ln x

C. dxxfx'f

u = f (x)

x'fdxdu

f'(x).dx = du

ln |f (x)| + C 1.

dx1x3x

3x22

2. dx

5ee

x

x

3. dxxcosxsin

1. u = x2 – 3 x + 1 2. u = ex + 5 3. u = cos x

D. dxxa,xf 22 x = a.Sin u

ucos.adudx

dx = a.cos u.du let op : √(a2 – x2) =a.cosu

1. dxxr 22

2. dxxr.x 22

3. 2x4

dx

1. x = r.sin u 2. x = r.sin u 3. x = 2.sin u