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How to integrate this => INT (3e^(2x) + 2/(25+x^2) dx?
INT 3e^(2x) + 2/(25+x^2) dx
= 3 e^(2x)/2 + 2 [ (1/5) * arctan (x/5) ] + C
= (3 e^(2x))/2 + [(2 * arctan(x/5) )/ 5 ] + C <====
1 Answer
- Anonymous1 decade agoFavorite Answer
first Part
∫3e^(2x) dx
u = 2x
du = 2 dx
dx = (1/2) du
∫3e^(2x) dx = ∫3e^u (1/2) du
= (3/2) e^u = (3/2) e^(2x)
Second part
∫ 2 dx /(25+x^2)
x = 5 tan(u)
dx = 5 sec^2(u) du
u = arctan(x/5)
∫ 2 dx /(25+x^2) = 2∫ 5 sec^2(u) du / (25 + 25 tan^2(u))
= (2/5)∫ sec^2(u) du / (1 + tan^2(u))
=(2/5)∫ sec^2(u) du / (sec^2(u))
= (2/5)∫ du
= 2u / 5 = (2/5)arctan(x/5)
Finally
add the two results together , and don't forget the constant
(3/2) e^(2x) + (2/5)arctan(x/5) + c