∫ e^x dx = ?
A
x*e^x + C
B
e^x + C
C
ln(e^x) + C
D
1/e^x + C
Analysis & Theory
The derivative of e^x is itself, so ∫ e^x dx = e^x + C.
∫ a^x dx (a > 0, a ≠ 1) = ?
A
a^x + C
B
a^x ln(a) + C
C
a^x/ln(a) + C
D
ln|x| + C
Analysis & Theory
∫ a^x dx = a^x / ln(a) + C.
∫ e^(2x) dx = ?
A
2e^(2x) + C
B
e^(2x)/2 + C
C
ln(e^(2x)) + C
D
e^(x) + C
Analysis & Theory
∫ e^(kx) dx = (1/k) e^(kx) + C. Here k = 2, so result = e^(2x)/2 + C.
∫ e^(-x) dx = ?
A
-e^(-x) + C
B
e^(-x) + C
C
ln|x| + C
D
1/e^x + C
Analysis & Theory
∫ e^(-x) dx = (1/-1) e^(-x) + C = -e^(-x) + C.
∫ x e^x dx = ?
A
x e^x - e^x + C
B
x e^x + C
C
e^x + C
D
x^2 e^x + C
Analysis & Theory
Use integration by parts: ∫ x e^x dx = x e^x - ∫ e^x dx = x e^x - e^x + C.
∫ (2x+1) e^(x^2+x) dx = ?
A
e^(x^2+x) + C
B
(2x+1)e^(x^2+x) + C
C
ln(e^(x^2+x)) + C
D
None of these
Analysis & Theory
Let u = x^2 + x ⇒ du = (2x+1) dx. So integral = ∫ e^u du = e^u + C = e^(x^2+x) + C.
∫ e^(3x+2) dx = ?
A
(1/3) e^(3x+2) + C
B
e^(3x+2) + C
C
ln(e^(3x+2)) + C
D
3 e^(3x+2) + C
Analysis & Theory
Let u = 3x+2 ⇒ du = 3 dx. So integral = (1/3) e^(3x+2) + C.
∫ (1/x) e^(ln x) dx = ?
A
e^(ln x) + C
B
x + C
C
ln|x| + C
D
1 + C
Analysis & Theory
e^(ln x) = x. So integrand = (1/x)·x = 1. ∫ 1 dx = x + C.
∫ e^(x^2) dx = ?
A
No elementary form
B
2e^(x^2) + C
C
e^(x^2)/2 + C
D
x e^(x^2) + C
Analysis & Theory
∫ e^(x^2) dx cannot be expressed in terms of elementary functions. It’s related to the error function.
∫ (ln x) e^(ln x) dx = ?
A
x ln x - x + C
B
x ln x + C
C
e^x + C
D
ln(x^2) + C
Analysis & Theory
e^(ln x) = x. So ∫ (ln x)·x dx. Use integration by parts: result = x ln x - x + C.