We will use integration by substitution, as well as the integrals ∫ 1 x d x = ln | x | + C and ∫ 1 d x = x + C ∫ x 3 x 2 + 1 d x = ∫ x 2 x 2 + 1 x d x = 1 2 ∫ ( x 2 + 1 ) − 1 x 2 + 1 2 x d x Let u = x 2 + 1 ⇒ d u = 2 x d x . Then 1 2 ∫ ( x 2 + 1 ) − 1 x 2 + 1 2 x d x = 1 2 ∫ u − 1 u d u = 1 2 ∫ ( 1 − 1 u ) d u = 1 2 ( u − ln | u | ) + C = x 2 + 1 2 − ln ( x 2 + 1 ) 2 + C = x 2 2 − ln ( x 2 + 1 ) 2 + 1 2 + C = x 2 − ln ( x 2 + 1 ) 2 + C Final answer
