Evaluate ∫ d x x 2 + 2 x + 2

Let I = ∫ d x x 2 + 2 x + 2 I = ∫ d x x 2 + 2 x + 1 2 - 1 2 + 2 I = ∫ d x x + 1 2 + 1 2 I = log | x + 1 + x + 1 2 + 1 2 | + c I = log | ( x + 1 ) + x 2 + 2 x + 1 | + c

Evaluate ∫ d x x 2 + 2 x + 2

Let I = ∫ d x x 2 + 2 x + 2 I = ∫ d x x 2 + 2 x + 1 2 - 1 2 + 2 I = ∫ d x x + 1 2 + 1 2 I = log | x + 1 + x + 1 2 + 1 2 | + c I = log | ( x + 1 ) + x 2 + 2 x + 1 | + c

Evaluate ∫ d x ( x - 1 ) ( x - 2 )

Let I = ∫ d x ( x - 1 ) ( x - 2 ) I = ∫ d x x 2 - 3 x + 2 I = ∫ d x x 2 - 3 x + 3 2 2 - 3 2 2 + 2 I = ∫ d x x - 3 2 2 - 1 2 2 I = log | x - 3 2 + x - 3 2 2 - 1 2 2 | I = log | x - 3 2 + x 2 - 3 x + 2 | + c

Evaluate ∫ d x 8 + 3 x - x 2

Let I = ∫ d x 8 + 3 x - x 2 I = ∫ d x - x 2 - 3 x - 8 = ∫ d x - x 2 - 3 x + 3 2 2 - 3 2 2 - 8 = ∫ d x - x - 3 2 2 - 41 2 2 = ∫ d x 41 2 2 - x - 3 2 2 = sin - 1 x - 3 2 41 2 + c I = sin - 1 2 x - 3 41 + c

Evaluate ∫ d x 5 x 2 - 2 x

Let I = ∫ d x 5 x 2 - 2 x I = ∫ d x 5 x 2 - 2 5 x = 1 5 ∫ d x x 2 - 2 x 5 + 1 5 2 - 1 5 2 = 1 5 ∫ d x x - 1 5 2 - 1 5 2 = 1 5 log | x - 1 5 + x - 1 5 2 - 1 5 2 | + c = 1 5 log | x - 1 5 + x 2 - 2 x 5 | + c = 1 5 log | 5 x - 1 5 + 5 x 2 - 2 x 5 | + c

Evaluate ∫ d x 9 x - 4 x 2

Let I = ∫ d x 9 x - 4 x 2 I = ∫ d x - 4 x 2 - 9 4 x = 1 2 ∫ d x - x 2 - 9 4 x + 9 16 - 9 16 = 1 2 ∫ d x 3 4 2 - x - 3 4 2 = 1 2 sin - 1 x - 3 4 3 4 + c I = 1 2 sin - 1 4 x - 3 3 + c

Evaluate ∫ d x 5 - 4 x - 2 x 2

Let I = ∫ d x 5 - 4 x - 2 x 2 I = ∫ d x - 2 x 2 + 2 x - 5 2 = 1 2 ∫ d x - x 2 + 2 x + 1 2 - 1 2 - 5 2 = 1 2 ∫ d x 7 2 - x + 1 2 = 1 2 ∫ d x 7 2 2 - x + 1 2 = 1 2 sin - 1 x + 1 7 2 + c I = 1 2 sin - 1 2 x + 1 7 + c

Evaluate ∫ x - 1 x 2 - x d x

Let I = ∫ x - 1 x 2 - x d x I = ∫ 1 2 2 x - 1 - 1 2 x 2 - x d x I = 1 2 ∫ 2 x - 1 x 2 - x d x - 1 2 ∫ d x x 2 - x For first integral let x 2 - x = t ( 2 x - 1 ) d x = d t I = 1 2 ∫ d t t - 1 2 ∫ d x x 2 - x + 1 2 2 - 1 2 2 I = 1 2 ∫ t - 1 2 d t - 1 2 ∫ d x x - 1 2 2 - 1 2 2 I = 1 2 t 1 2 1 2 - 1 2 log | x - 1 2 + x - 1 2 2 - 1 2 2 I = t - 1 2 log | x - 1 2 + x 2 - x | + c I = x 2 - x - 1 2 log | x - 1 2 + x 2 - x | + c

Evaluate ∫ x + 3 5 - 4 x - x 2 d x

Let I = ∫ x + 3 5 - 4 x - x 2 d x I = ∫ - 1 2 - 4 - 2 x + 1 5 - 4 x - x 2 d x I = - 1 2 ∫ - 4 - 2 x 5 - 4 x - x 2 d x + ∫ 1 5 - 4 x - x 2 d x For first Integral 5 - 4 x - x 2 = t - 4 - 2 x d x = d t I = - 1 2 ∫ d t t + ∫ d x - x 2 + 4 x - 5 I = - 1 2 ∫ t - 1 2 d t + ∫ d x - x 2 + 4 x + 2 2 - 2 2 - 5 I = - 1 2 t 1 2 1 2 + ∫ d x 3 2 - x + 2 2 I = - 5 - 4 x - x 2 + sin - 1 x + 2 3 + c

Evaluate ∫ 5 x + 3 x 2 + 4 x + 10 d x

Let I = ∫ 5 x + 3 x 2 + 4 x + 10 d x I = ∫ 5 2 2 x + 4 - 7 x 2 + 4 x + 10 d x I = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 d x - 7 ∫ 1 x 2 + 4 x + 10 d x For first integral Substitute x 2 + 4 x + 10 = t Differentiate with respect to x , then ( 2 x + 4 ) d x = d t I = 5 2 ∫ d t t - 7 ∫ d x x 2 + 4 x + 2 2 - 2 2 + 10 I = 5 2 ∫ t - 1 2 d t - 7 ∫ d x x + 2 2 + 6 2 I = 5 2 t 1 2 1 2 - 7 log ( x + 2 ) + x + 2 2 + 6 2 + C = 5 x 2 + 4 x + 10 - 7 log x + 2 + x 2 + 4 x + 10 + C