Problem: ∫ x ln ( x + 1 ) d x ∫xln(x+1)dx Integrate by parts: ∫ f g ′ = f g − ∫ f ′ g ∫fg′=fg−∫f′g f f = ln ( x + 1 ) =ln(x+1), g ′ g′ = x =x ↓ ↓ Steps ↓ ↓ Steps f ′ f′ = 1 x + 1 =1x+1, g g = x 2 2 =x22: = x 2 ln ( x + 1 ) 2 − ∫ x 2 2 ( x + 1 ) d x =x2ln(x+1)2−∫x22(x+1)dx Now solving: ∫ x 2 2 ( x + 1 ) d x ∫x22(x+1)dx Apply linearity: = 1 2 ∫ x 2 x + 1 d x =12∫x2x+1dx Now solving: ∫ x 2 x + 1 d x ∫x2x+1dx Substitute u = x + 1 u=x+1 ⟶ ⟶ d u d x = 1 dudx=1 (Steps), use x 2 = ( u − 1 ) 2 x2=(u−1)2: = ∫ ( u − 1 ) 2 u d u =∫(u−1)2udu … or choose an alternative: Substitute 1 x + 1 1x+1 Don't substitute Expand: = ∫ ( u + 1 u − 2 ) d u =∫(u+1u−2)du Apply linearity: = ∫ u d u + ∫ 1 u d u − 2 ∫ 1 d u =∫udu+∫1udu−2∫1du Now solving: ∫ u d u ∫udu Apply power rule: ∫ u n d u = u n + 1 n + 1 ∫undu=un+1n+1 with n = 1 n=1: = u 2 2 =u22 Now solving: ∫ 1 u d u ∫1udu This is a standard integral: = ln ( u ) =ln(u) Now solving: ∫ 1 d u ∫1du Apply constant rule: = u =u Plug in solved integrals: ∫ u d u + ∫ 1 u d u − 2 ∫ 1 d u ∫udu+∫1udu−2∫1du = ln ( u ) + u 2 2 − 2 u =ln(u)+u22−2u Undo substitution u = x + 1 u=x+1: = ln ( x + 1 ) + ( x + 1 ) 2 2 − 2 ( x + 1 ) =ln(x+1)+(x+1)22−2(x+1) Plug in solved integrals: 1 2 ∫ x 2 x + 1 d x 12∫x2x+1dx = ln ( x + 1 ) 2 + ( x + 1 ) 2 4 − x − 1 =ln(x+1)2+(x+1)24−x−1 Plug in solved integrals: x 2 ln ( x + 1 ) 2 − ∫ x 2 2 ( x + 1 ) d x x2ln(x+1)2−∫x22(x+1)dx = x 2 ln ( x + 1 ) 2 − ln ( x + 1 ) 2 − ( x + 1 ) 2 4 + x + 1 =x2ln(x+1)2−ln(x+1)2−(x+1)24+x+1 The problem is solved: ∫ x ln ( x + 1 ) d x ∫xln(x+1)dx = x 2 ln ( x + 1 ) 2 − ln ( x + 1 ) 2 − ( x + 1 ) 2 4 + x + 1 + C =x2ln(x+1)2−ln(x+1)2−(x+1)24+x+1+C Rewrite/simplify: = ( x + 1 ) ( 2 ( x − 1 ) ln ( x + 1 ) − x + 3 ) 4