Cracking The Ap Calculus ab Exam 2018

we  need    to  substitute  for x   dx, so  if  we  divide  du  by  10, we  get  du =   x   dx. Now,    if  we

substitute into the integrand, we get ∫ xe^5 x

(^2) −1
dx = (^) ∫ eu du = eu + C. Substituting back,

we get ∫ xe^5 x

(^2) −1
dx = e^5 x
(^2) −1

• C.

8. ln |ex − e−x| + C

Recall that ∫eudu = eu + C. Let’s use u-substitution. If we let u = ex − e−x, then du = ex + e−xdx.

If we substitute into the integrand, we get ∫ dx = ∫ . Recall that ∫ = ln |u | + C.

Substituting back, we get ∫ dx = ln | ex − e−x| + C.

9.

Recall that ∫ au du = au + C. Let’s use u-substitution. If we let u = −x^2 , then du = −2x dx.

But we  need    to  substitute  for x   dx, so  if  we  divide  du  by  −2, we  get − du    =   x   dx. Now,    if  we

substitute into the integrand, we get ∫ x4−x2 dx = − (^) ∫ 4 u du = − 4 u + C. Substituting

back, we get ∫ x 4 −x2 dx = − 4 −x2 + C = − 4 −x2 + C.

10.

Recall that ∫ au du = au + C. Let’s use u-substitution. If we let u = sin x, then du = cos x

dx. If we substitute into the integrand, we get ∫ 7 sin x cos x dx = ∫ 7 u du = 7 u + C.

Substituting back, we get ∫ 7 sin x cos x dx = + C.

SOLUTIONS TO UNIT 3 DRILL

1.