Cracking The Ap Calculus ab Exam 2018

. Now we can integrate: = =

. Last, we substitute back and get .

4.

If  we  let u   =   x^3     +   3x, then    du  =   (3x^2   +   3)  dx. We  need    to  substitute  for (x^2    +   1)  dx, so  we  can

divide the du term by 3: = (x^2 + 1) dx. Next we can substitute into the integral: ∫(x^2 + 1)(x^3

• 3x)−5 dx = (^) ∫u−5 du. Now we can integrate: = . Last,
  we substitute back and get .

5. −2 cos + C

If  we  let u   =    ,  then    du  =    dx.    We  need    to  substitute  for  dx,    so  we  can multiply    the

du term by 2: 2du = dx. Next we can substitute into the integral: ∫ sin dx = 2∫ sin

u du. Now we can integrate: 2 ∫sin u du = −2 cos u + C. Last, we substitute back and get −2

cos     +   C.

6. tan (x^3 ) + C

If  we  let u   =   x^3 ,   then    du  =   3x^2 dx.    We  need    to  substitute  for x^2 dx, so  we  can divide  the du  term

by 3: = x^2 dx. Next we can substitute into the integral: ∫x^2 sec^2 (x^3 ) dx = (^) ∫ sec^2 u du.
Now we can integrate: (^) ∫ sec^2 u du = tan u + C. Last, we substitute back and get tan (x^3 )

+ C.