logarithm. A clue is whether the numerator is the derivative of the denominator, as it is in the
first integral. Let’s use u-substitution. If we let u = cos x, then du = −sin x dx. If we substitute
into the integrand, we get dx = − . Recall that = ln | u | + C. Substituting
back, we get dx = − ln|cos x | + C. The second integral is simply ∫ dx = x + C.
Therefore, the integral is dx − ∫ dx = − ln | cos x | − x + C.
- ln (1 + 2 ) + C
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
logarithm. A clue is whether the numerator is the derivative of the denominator, as it is here.
Let’s use u-substitution. If we let u = 1 + 2 , then du = dx. If we substitute into the
integrand, we get dx = ∫ . Recall that ∫ = ln | u | + C. Substituting back,
we get ∫ dx = ln (1 + 2 ) + C. Notice that we don’t need the absolute value
bars because 1 + 2 is never negative.
- ln (1 + ex) + C
Whenever we have an integral in the form of a quotient, we check to see if the solution is a
logarithm. A clue is whether the numerator is the derivative of the denominator, as it is here.
Let’s use u-substitution. If we let u = 1 + ex, then du = exdx. If we substitute into the integrand,
we get ∫ dx = ∫ . Recall that ∫ = ln | u | + C. Substituting back, we get ∫
dx = ln (1 + ex) + C. Notice that we don’t need the absolute value bars because 1 + ex is never
negative.
7.
Recall that ∫eudu = eu + C. Let’s use u-substitution. If we let u = 5x^2 − 1, then du = 10x dx. But
