Integrate the function 4 x + 1 2 x 2 + x - 3

4 x + 1 2 x 2 + x - 3 Let 2 x 2 + x - 3 = t Differentiate both sides w.r.t x 4 x + 1 + 0 = d t d x d x = d t 4 x + 1 Now, ∫ 4 x + 1 2 x 2 + x - 3 d x Put the values of 2 x 2 + x - 3 = t and d x = d t 4 x + 1 = ∫ 4 x + 1 t × d t 4 x + 1 = ∫ 1 t d t = ∫ t - 1 2 d t = t - 1 2 + 1 - 1 2 + 1 + C = t 1 2 1 2 + C = 2 t 1 2 + C = 2 t + C = 2 2 x 2 + x - 3 + C

Integrate the function x + 2 x 2 - 1

∫ x + 2 x 2 - 1 · d x = ∫ 1 2 ( 2 x ) + 2 x 2 - 1 · d x = ∫ 1 2 ( 2 x ) x 2 - 1 · d x + ∫ 2 x 2 - 1 · d x = 1 2 ∫ 2 x x 2 - 1 · d x + ∫ 2 x 2 - 1 · d x ...........(i) Let I 1 = 1 2 ∫ 2 x x 2 - 1 d x Let x 2 - 1 = t Differentiating w.r.t x 2 x - 0 = d t d x d x = d t 2 x Thus, our equation becomes I 1 = 1 2 ∫ 2 x x 2 - 1 d x Put the values of x 2 - 1 = t and d x = d t 2 x , we get I 1 = 1 2 ∫ 2 x t d x I 1 = 1 2 ∫ 2 x t × d t 2 x I 1 = 1 2 ∫ 1 t d t I 1 = 1 2 ∫ t - 1 2 d t I 1 = 1 2 t - 1 2 + 1 - 1 2 + 1 + C 1 I 1 = 1 2 t 1 2 1 2 + C 1 I 1 = t 1 2 + C 1 I 1 = t + C 1 I 1 = x 2 - 1 + C 1 (Using t = x 2 - 1 ) Let I 2 = ∫ 2 x 2 - 1 · d x I 2 = 2 ∫ 1 x 2 - ( 1 ) 2 · d x It is of form ∫ d x x 2 - a 2 = log | x + x 2 - a 2 | + C ∴ Replacing a by 1 we get I 2 = 2 log | x + x 2 - 1 | + C 2 Now, Putting the values of I 1 and I 2 in ( 1 ) ∫ x + 2 x 2 - 1 · d x = 1 2 ∫ 2 x x 2 - 1 · d x + 2 ∫ 1 x 2 - 1 · d x = x 2 - 1 + C 1 + 2 log | x + x 2 - 1 | + C 2 = x 2 - 1 + 2 log | x + x 2 - 1 | + c

Integrate the function 5 x - 2 1 + 2 x + 3 x 2

∫ 5 x - 2 1 + 2 x + 3 x 2 d x = 5 ∫ x - 2 5 1 + 2 x + 3 x 2 d x = 5 6 ∫ 6 x - 12 5 1 + 2 x + 3 x 2 d x = 5 6 ∫ 6 x + 2 - 12 5 - 2 1 + 2 x + 3 x 2 d x = 5 6 ∫ 6 x + 2 - 22 5 1 + 2 x + 3 x 2 d x = 5 6 ∫ 6 x + 2 1 + 2 x + 3 x 2 d x - 5 6 × 22 5 ∫ d x 1 + 2 x + 3 x 2 = 5 6 ∫ 6 x + 2 1 + 2 x + 3 x 2 d x - 11 3 ∫ d x 1 + 2 x + 3 x 2 . . . . . . . . . . 1 Let I 1 = 5 6 ∫ 2 + 6 x 3 x 2 + 2 x + 1 d x Let 3 x 2 + 2 x + 1 = t Differentiate both sides w.r.t x 6 x + 2 + 0 = d t d x d x = d t 6 x + 2 Thus, our equation becomes I 1 = 5 6 ∫ 2 + 6 x 3 x 2 + 2 x + 1 d x Put the values of 3 x 2 + 2 x + 1 = t and d x = d t 6 x + 2 , we get I 1 = 5 6 ∫ 6 x + 2 t d x I 1 = 5 6 ∫ 6 x + 2 t × d t 6 x + 2 I 1 = 5 6 ∫ 1 t · d t I 1 = 5 6 log | t | + C 1 I 1 = 5 6 log 3 x 2 + 2 x + 1 + C 1 (Using t = 3 x 2 + 2 x + 1 ) Let I 2 = 11 3 ∫ 1 3 x 2 + 2 x + 1 · d x = 11 3 ∫ 1 3 x 2 + 2 x 3 + 1 3 · d x (Taking 3 common) = 11 3 . 3 ∫ 1 x 2 + 2 x 3 + 1 3 · d x = 11 9 ∫ 1 x 2 + 2 ( x ) 1 3 + 1 3 · d x = 11 9 ∫ 1 x 2 + 2 ( x ) 1 3 + 1 3 · d x = 11 9 ∫ 1 x 2 + 2 ( x ) 1 3 + 1 3 2 - 1 3 2 + 1 3 · d x = 11 9 ∫ 1 x + 1 3 2 - 1 3 2 + 1 3 · d x = 11 9 ∫ 1 x + 1 3 2 - 1 9 + 1 3 · d x = 11 9 ∫ 1 x + 1 3 2 - 1 9 + 1 3 · d x = 11 9 ∫ 1 x + 1 3 2 + 2 9 · d x = 11 9 ∫ 1 x + 1 3 2 + 2 3 2 · d x It is of form ∫ 1 x 2 + a 2 d x = 1 a tan - 1 x a + C 2 ∴ Replacing x by x + 1 3 and a by 2 3 , we get = 11 9 1 2 3 tan - 1 x + 1 3 2 3 + C 2 = 11 9 3 2 tan - 1 3 x + 1 3 2 3 + C 2 = 11 9 3 2 tan - 1 3 x + 1 2 = 11 9 · 3 2 tan - 1 3 x + 1 2 + 11 9 C 2 = 11 3 2 tan - 1 3 x + 1 2 + C 3 (Where C 3 = 11 9 C 2 ) Now, putting the values of I 1 and I 2 in ( 1 ) ∫ 5 x - 2 1 + 2 x + 3 x 2 · d x = 5 6 ∫ 2 + 6 x 1 + 2 x + 3 x 2 · d x - 11 3 ∫ 1 1 + 2 x + 3 x 2 · d x = 5 6 log 3 x 2 + 2 x + 1 - 11 3 2 tan - 1 3 x + 1 2 + c

Integrate 6 x + 7 ( x - 5 ) ( x - 4 )

∫ 6 x + 7 ( x - 5 ) ( x - 4 ) · d x = ∫ 6 x + 7 x ( x - 4 ) - 5 ( x - 4 ) · d x = ∫ 6 x + 7 x 2 - 4 x - 5 x + 20 · d x = ∫ 6 x + 7 x 2 - 9 x + 20 · d x = 6 ∫ x + 7 6 x 2 - 9 x + 20 · d x = 6 2 ∫ 2 x + 14 6 x 2 - 9 x + 20 · d x = 3 ∫ 2 x - 9 + 7 3 + 9 x 2 - 9 x + 20 · d x = 3 ∫ 2 x - 9 + 34 3 x 2 - 9 x + 20 · d x = 3 ∫ 2 x - 9 + 34 3 x 2 - 9 x + 20 · d x = 3 ∫ 2 x - 9 x 2 - 9 x + 20 d x + 3 × 34 3 ∫ d x x 2 - 9 x + 20 = 3 ∫ 2 x - 9 x 2 - 9 x + 20 d x + 34 ∫ d x x 2 - 9 x + 20 . . . . . . . . . . 1 Let I 1 = 3 ∫ 2 x - 9 x 2 - 9 x + 20 · d x I 1 = 3 ∫ 2 x - 9 x 2 - 9 x + 20 · d x Let, x 2 - 9 x + 20 = t Differentiate both sides w.r.t x 2 x - 9 + 0 = d t d x d x = d t 2 x - 9 Thus, our equation becomes I 1 = 3 ∫ 2 x - 9 x 2 - 9 x + 20 · d x Putting values of t = x 2 - 9 x + 20 and d x = d t 2 x - 9 , we get I 1 = 3 ∫ 2 x - 9 t · d x I 1 = 3 ∫ 2 x - 9 t · d t 2 x - 9 I 1 = 3 ∫ 1 t · d t I 1 = 3 ∫ 1 ( t ) 1 2 · d t I 1 = 3 ∫ ( t ) - 1 2 · d t I 1 = 3 t - 1 2 + 1 - 1 2 + 1 + C 1 I 1 = 3 t 1 2 1 2 + C 1 I 1 = 3 . 2 t 1 2 + C 1 I 1 = 6 t 1 2 + C 1 I 1 = 6 t 1 2 + C 1 I 1 = 6 t + C 1 I 1 = 6 x 2 - 9 x + 20 + C 1 Let I 2 = 34 ∫ 1 x 2 - 9 x + 20 · d x I 2 = 34 ∫ 1 x 2 - 2 ( x ) 9 2 + 20 · d x I 2 = 34 ∫ 1 x 2 - 2 ( x ) 9 2 + 9 2 2 - 9 2 2 + 20 · d x I 2 = 34 ∫ 1 x - 9 2 2 - 9 2 2 + 20 · d x I 2 = 34 ∫ 1 x - 9 2 2 - 81 4 + 20 · d x I 2 = 34 ∫ 1 x - 9 2 2 - 1 4 · d x I 2 = 34 ∫ 1 x - 9 2 2 - 1 2 2 · d x It is of form ∫ d x x 2 - a 2 = log | x + x 2 - a 2 | + C ∴ Replacing x by x - 9 2 and a by 1 2 , we get = 34 log x - 9 2 + x - 9 2 2 + 1 2 2 + C 2 = 34 log x - 9 2 + x 2 + 9 2 2 - 2 ( x ) 9 2 - 1 4 + C 2 = 34 log x - 9 2 + x 2 + 81 4 - 9 x - 1 4 + C 2 = 34 log x - 9 2 + x 2 - 9 x + 81 4 - 1 4 + C 2 = 34 log x - 9 2 + x 2 - 9 x + 80 4 + C 2 = 34 log x - 9 2 + x 2 - 9 x + 20 + C 2 Now, putting values of I 1 and I 2 in equation ( 1 ) ∫ 6 x + 7 ( x - 2 ) ( x - 4 ) · d x = I 1 + I 2 = 6 x 2 - 9 x + 20 + C 1 + 34 log x - 9 2 + x 2 - 9 x + 20 + C = 6 x 2 - 9 x + 20 + 34 log x - 9 2 + x 2 - 9 x + 20 + C

Integrate the function x + 2 4 x - x 2

∫ x + 2 4 x - x 2 . d ( x ) = - 1 2 ∫ - 2 x - 4 4 x - x 2 d ( x ) = - 1 2 ∫ - 2 x + 4 - 4 - 4 4 x - x 2 d x = - 1 2 ∫ - 2 x + 4 - 8 4 x - x 2 d x = - 1 2 ∫ - 2 x + 4 4 x - x 2 d x + 8 2 ∫ d x 4 x - x 2 = - 1 2 ∫ - 2 x + 4 4 x - x 2 d x + 4 ∫ d x 4 x - x 2 . . . . . . . . . 1 Let I 1 = - 1 2 ∫ 4 - 2 x 4 x - x 2 · d x Let 4 x - x 2 = t Differentiate both sides w.r.t. x 4 - 2 x = d t d x d x = d t 4 - 2 x Thus, our equation becomes I 1 = - 1 2 ∫ 4 - 2 x 4 x - x 2 · d x Putting the values of 4 x - x 2 and d x ∴ I 1 = - 1 2 ∫ 4 - 2 x t · d x I 1 = - 1 2 ∫ 4 - 2 x t · d t 4 - 2 x I 1 = - 1 2 ∫ 1 t d t I 1 = - 1 2 ∫ 1 ( t ) 1 2 d t I 1 = - 1 2 t - 1 2 + 1 - 1 2 + 1 + C 1 I 1 = - 1 2 t 1 2 1 2 + C 1 I 1 = - t 1 2 + C 1 I 1 = - t + C 1 I 1 = - 4 x - x 2 + C 1 (Using t = 4 x - x 2 ) Let I 2 = 4 ∫ 1 4 x - x 2 · d x I 2 = 4 ∫ 1 - x 2 - 4 x · d x I 2 = 4 ∫ 1 - x 2 - 2 ( 2 ) ( x ) · d x I 2 = 4 ∫ 1 - x 2 - 2 ( 2 ) ( x ) + ( 2 ) 2 - ( 2 ) 2 · d x I 2 = 4 ∫ 1 - ( x - 2 ) 2 - ( 2 ) 2 · d x I 2 = 4 ∫ 1 - ( x - 2 ) 2 - 4 · d x I 2 = 4 ∫ 1 4 - ( x - 2 ) 2 · d x I 2 = 4 ∫ 1 ( 2 ) 2 - ( x - 2 ) 2 · d x It is of form ∫ d x a 2 - x 2 = sin - 1 x a + C 1 ∴ Replacing x by ( x - 2 ) and a by 2 we get I 2 = 4 ∫ 1 ( 2 ) 2 - ( x - 2 ) 2 · d x I 2 = 4 sin - 1 x - 2 2 + C 2 Putting values of I 1 and I 2 in equation ( 1 ) ∫ x + 2 4 x - x 2 · d x = - 1 2 ∫ 4 - 2 x 4 x - x 2 · d x + 4 ∫ 1 4 x - x 2 · d x = - 4 x - x 2 + C 1 + 4 sin - 1 x - 2 2 + C 2 = - 4 x - x 2 + 4 sin - 1 x - 2 2 + C

Integrate the function x + 2 x 2 + 2 x + 3

∫ x + 2 x 2 + 2 x + 3 d x = 1 2 ∫ 2 x + 4 x 2 + 2 x + 3 d x = 1 2 ∫ 2 x + 2 + 4 - 2 x 2 + 2 x + 3 d x = 1 2 ∫ 2 x + 2 x 2 + 2 x + 3 d x + 2 2 ∫ d x x 2 + 2 x + 3 d x = 1 2 ∫ 2 x + 2 x 2 + 2 x + 3 d x + ∫ d x x 2 + 2 x + 3 d x . . . . . . . . 1 let I 1 = 1 2 ∫ 2 x + 2 x 2 + 2 x + 3 · d x Let, x 2 + 2 x + 3 = t Differentiate both sides w.r.t x 2 x + 2 + 0 = d t d x d x = d t 2 x + 2 Now, our equation becomes I 1 = 1 2 ∫ 2 x + 2 x 2 + 2 x + 3 · d x Putting the value of 4 x - x 2 and d x I 1 = 1 2 ∫ 2 x + 2 t · d x I 1 = 1 2 ∫ 2 x + 2 t · d t 2 x + 2 I 1 = 1 2 ∫ 1 t · d t I 1 = 1 2 ∫ 1 ( t ) 1 2 · d t I 1 = 1 2 ∫ ( t ) - 1 2 · d t I 1 = 1 2 t - 1 2 + 1 - 1 2 + 1 + C 1 I 1 = t 1 2 + C 1 I 1 = t + C 1 I 1 = x 2 + 2 x + 3 + C (Using t = x 2 + 2 x + 3 ) Let I 2 = ∫ 1 x 2 + 2 x + 3 · d x I 2 = ∫ 1 x 2 + 2 ( x ) ( 1 ) + 3 · d x Adding and subtracting 1 2 I 2 = ∫ 1 x 2 + 2 ( x ) ( 1 ) - ( 1 ) 2 + ( 1 ) 2 + 3 · d x I 2 = ∫ 1 ( x + 1 ) 2 - ( 1 ) 2 + 3 · d x I 2 = ∫ 1 ( x + 1 ) 2 - 1 + 3 · d x I 2 = ∫ 1 ( x + 1 ) 2 + 2 · d x I 2 = ∫ 1 ( x + 1 ) 2 + ( 2 ) 2 · d x It is of form ∫ d x x 2 + a 2 = log | x + x 2 + a 2 | + C 2 ∴ Replacing x by ( x + 1 ) and a by 2 , we get I 2 = log | x + 1 + ( x + 1 ) 2 + ( 2 ) 2 | + C 2 I 2 = log | x + 1 + x 2 + 2 x + 1 + 2 | + C 2 I 2 = log | x + 1 + x 2 + 2 x + 3 | + C 2 Putting the values of I 1 and I 2 in ( 1 ) ∫ x + 2 x 2 + 2 x + 3 · d x = I 1 + I 2 = x 2 + 2 x + 3 + C 1 + log | x + 1 + x 2 + 2 x + 3 | + C 2 = x 2 + 2 x + 3 + log | x + 1 + x 2 + 2 x + 3 | + c

Integrate the function x + 3 x 2 - 2 x - 5

∫ x + 3 x 2 - 2 x - 5 d x = 1 2 ∫ 2 x + 6 x 2 - 2 x - 5 d x = 1 2 ∫ 2 x - 2 + 2 + 6 x 2 - 2 x - 5 d x = 1 2 ∫ 2 x - 2 x 2 - 2 x - 5 d x + 8 2 ∫ d x x 2 - 2 x - 5 = 1 2 ∫ 2 x - 2 x 2 - 2 x - 5 d x + 4 ∫ d x x 2 - 2 x - 5 . . . . . . . . . . . 1 Let I 1 = 1 2 ∫ 2 x - 2 x 2 - 2 x - 5 · d x Let x 2 - 2 x - 5 = t Differentiate both sides w.r.t x we get, 2 x - 2 - 0 = d t d x d x = d t 2 x - 2 Thus, our equation becomes ∴ I 1 = 1 2 ∫ 2 x - 2 x 2 - 2 x - 5 · d x Putting value of x 2 - 2 x - 5 = t and d x 2 x - 2 = d t ⇒ d x = d t 2 x - 2 I 1 = 1 2 ∫ 2 x - 2 t · d x I 1 = 1 2 ∫ 2 x - 2 t · d t 2 x - 2 I 1 = 1 2 ∫ 1 t · d t I 1 = 1 2 log | t | + C 1 (Using t = x 2 - 2 x - 5 ) I 1 = 1 2 log x 2 - 2 x - 5 + C 1 Let I 2 = 4 ∫ 1 x 2 - 2 x - 5 · d x I 2 = 4 ∫ 1 x 2 - 2 ( x ) ( 1 ) - 5 · d x I 2 = 4 ∫ 1 x 2 - 2 ( x ) ( 1 ) + ( 1 ) 2 - ( 1 ) 2 - 5 · d x I 2 = 4 ∫ 1 ( x - 1 ) 2 - ( 1 ) 2 - 5 I 2 = 4 ∫ 1 ( x - 1 ) 2 - 1 - 5 · d x I 2 = 4 ∫ 1 ( x - 1 ) 2 - 6 · d x I 2 = 4 ∫ 1 ( x - 1 ) 2 - ( 6 ) 2 · d x It is of the form: ∫ d x x 2 - a 2 = 1 2 a log x - a x + a + C ∴ Replacing x by ( x - 1 ) and a by 6 , we get I 2 = 4 2 6 log x - 1 - 6 x - 1 + 6 + C 2 I 2 = 2 6 log x - 1 - 6 x - 1 + 6 + C 2 Therefore, ⇒ ∫ x + 3 x 2 - 2 x - 5 · d x = I 1 + I 2 = 1 2 log x 2 - 2 x - 5 + C 1 + 2 6 log x - 1 - 6 x - 1 + 6 + C 2 = 1 2 log x 2 - 2 x - 5 + 2 6 log x - 1 - 6 x - 1 + 6 + C

Integrate the function 5 x + 3 x 2 + 4 x + 10

∫ 5 x + 3 x 2 + 4 x + 10 d x = 5 ∫ x + 3 5 x 2 + 4 x + 10 = 5 2 ∫ 2 x + 6 5 x 2 + 4 x + 10 = 5 2 ∫ 2 x + 4 - 4 + 6 5 x 2 + 4 x + 10 = 5 2 ∫ 2 x + 4 - 14 5 x 2 + 4 x + 10 = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 d x + 5 2 ∫ - 14 5 x 2 + 4 x + 10 d x = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 d x - 7 ∫ 1 x 2 + 4 x + 10 d x . . . . . . . . . . . . . . 1 Solving I 1 I 1 = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 · d x Let x 2 + 4 x + 10 = t Differentiate both sides w.r.t. x 2 x + 4 + 0 = d t d x d x = d t 2 x + 4 Thus, our equation becomes I 1 = 5 2 ∫ 2 x + 4 x 2 + 4 x + 10 · d x Putting the value of x 2 + 4 x + 10 = t and d x = d t 2 x + 4 I 1 = 5 2 ∫ 2 x + 4 t · d x I 1 = 5 2 ∫ 2 x + 4 t · d t 2 x + 4 I 1 = 5 2 ∫ 1 t · d t I 1 = 5 2 ∫ 1 ( t ) 1 2 · d t I 1 = 5 2 ∫ ( t ) - 1 2 · d t I 1 = 5 2 t - 1 2 + 1 - 1 2 + 1 + C 1 I 1 = 5 2 t 1 2 1 2 + C 1 I 1 = 5 t 1 2 + C 1 I 1 = 5 t + C 1 I 1 = 5 x 2 + 4 x + 10 + C 1 (Using t = x 2 + 4 x + 1 ) Let I 2 = ∫ 7 x 2 + 4 x + 10 · d x I 2 = 7 ∫ 1 x 2 + 2 ( 2 ) ( x ) + 10 · d x I 2 = 7 ∫ 1 x 2 + 2 ( 2 ) ( x ) + ( 2 ) 2 - ( 2 ) 2 + 10 · d x I 2 = 7 ∫ 1 ( x + 2 ) 2 - ( 2 ) 2 + 10 · d x I 2 = 7 ∫ 1 ( x + 2 ) 2 - 4 + 10 · d x I 2 = 7 ∫ 1 ( x + 2 ) 2 + 6 · d x I 2 = 7 ∫ 1 ( x + 2 ) 2 + ( 6 ) 2 · d x It is of form ∫ d x x 2 + a 2 = log | x + x 2 + a 2 | + C 2 ∴ Replacing x by ( x + 2 ) and a by 6 , we get I 2 = 7 [ log | x + 2 + ( x + 2 ) 2 + ( 6 ) 2 | ] + C 2 I 2 = 7 log | x + 2 + x 2 + 4 x + 4 + 6 | + C 2 I 2 = 7 log | x + 2 + x 2 + 4 x + 10 | + C 2 Putting the values of I 1 and I 2 in ( 1 ) ∫ 5 x + 3 x 2 + 4 x + 10 · d x = I 1 - I 2 = 5 x 2 + 4 x + 10 + C 1 - 7 log | x + 2 + x 2 + 4 x + 10 | + C 2 = 5 x 2 + 4 x + 10 - 7 log | x + 2 + x 2 + 4 x + 10 | + C

E M B I B E

MATHEMATICS PART II : Class XII>Chapter 1 - Integrals>EXERCISE

EXERCISE

Author:Jharkhand Board

Jharkhand Board Mathematics Solutions for EXERCISE

Simple step-by-step solutions to EXERCISE questions of Integrals from MATHEMATICS PART II : Class XII. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.