3.
We find the derivative of the inverse tangent using the formula (tan−1 u) = .
Here we have u = ex, so = ex. Therefore, (tan−1ex) = .
4.
Recall that = sec−1 u + C. Here we have , and we just need to
rearrange the integrand so that it is in the proper form to use the integral formula. If we factor π
out of the radicand, we get . Next, we do u-substitution. Let u = and du
= dx. Multiply both by so that u = x and du = dx. Substituting into the
integrand, we get = . Now we get
sec−1 u + C. Substituting back, we get .
5.
Recall that = tan−1 u + C. Here we have , and we just need to rearrange the
integrand so that it is in the proper form to use the integral formula. If we factor π out of the
denominator, we get = . Next, we do u-substitution. Let u
= and du = dx. Multiply du by so that du = dx. Substituting into the
integrand, we get = . Now we get = tan−1 u + C.
