integral of 5e^x-cos(x)+3x^2-1/x
1) Let's compute the integral: ∫(5ex -cos(x) + 3x2 -1x)dx
For the math expression: -cos(x), since (sin(a))' = a'*cos(a), we try differentiating sin(x)
2) Let's perform the differentiation: ddx(sin(x))
ⓘ The derivative of sin(x) is cos(x): ddx(sin(x)) = cos(x)
cos(x)
You observe that the computed derivative and the original math expression are the same, except for a coefficient. So the integral is:
ⅰ ∫(5ex -cos(x) + 3x2 -1x)dx = -sin(x) + ∫(5ex + 3x2 -1x)dx
-sin(x) + ∫(5ex + 3x2 -1x)dx
For the math expression: 5ex, since (exp(a))' = a'*exp(a), we try differentiating exp(x)
3) Let's perform the differentiation: ddx(exp(x))
ⓘ The derivative of exp(x) is exp(x): ddx(exp(x)) = exp(x)
exp(x)
The computed derivative matches the original math expression up to a constant factor. Therefore, the integral is:
ⅰ ∫(5ex + 3x2 -1x)dx = 5ex + ∫(3x2 -1x)dx
-sin(x) + (5ex + ∫(3x2 -1x)dx)
ⅰ ∫(3x2 -1x)dx = ∫3x2dx -∫1xdx
-sin(x) + 5ex + (∫3x2dx -∫1xdx)
ⓘ A constant factor stays in front of the integral. An antiderivative of x^n is x^(n+1)/(n+1), for n ≠ -1: ∫3x2dx = 3·x33■The derivative of -ln|x| is -1x, so -ln|x| is an antiderivative of -1x: -∫1xdx = -ln|x|
-sin(x) + 5ex + 3·x33 -ln|x|
ⓘ Multiplying a fraction by a number multiplies its numerator: 3·x33 = 3x33■A fraction is simplified by dividing the numerator and the denominator by the same number: 3x33 = x3
-sin(x) + 5ex + x3 -ln|x|
▶Indefinite integrals are defined up to an additive constant C, so this yields: 5ex + x3 -ln|x| -sin(x) + C