Antiderivatives such as ∫ d x e 2 x + c

Caleb Proctor

Caleb Proctor

Answered

2022-07-04

Antiderivatives such as d x e 2 x + c

Answer & Explanation

Jamarcus Shields

Jamarcus Shields

Expert

2022-07-05Added 17 answers

Step 1
Firstly, the integral 1 u 2 c 2 d u = 1 c artanh u c + C is quite standard, so really if you'd have known it your method would be pretty efficient!
Step 2
Secondly, you could first make the substitution u = e x / c , so that e x = u c and d u / d x = e x / c = u d x = d u / u. This means that if I is the integral then I = 1 e 2 x + c d x = 1 u 2 c + c × 1 u d u = 1 c 1 u u 2 + 1 d u
Here we have a not-very-standard integral: 1 u u 2 + 1 d u = arcosech u + C = arsinh 1 u + C
ntaraxq

ntaraxq

Expert

2022-07-06Added 3 answers

Step 1
You can write this integral as I = e x d x 1 + c e 2 x
and make the substitution (for c > 0):
c e x = sinh t
1 + c e 2 x = cosh t
c e x d x = cosh t d t
Step 2
Then you immediately get
I = 1 c d t = 1 c a r s i n h ( c e x ) + c o n s t ..
For c < 0 there's a similar substitution, c e x = sin t

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