Obtain the differential coefficient of the following (i) x – 1 2 x + 5 (ii) 1 2 x + 1 (iii) 3 x + 4 4 x + 5 (iv) x 2 x 3 + 1

(i) d d x ( x - 1 ) ( 2 x + 5 ) Using differentiating by parts method: = ( x - 1 ) d d x ( 2 x + 5 ) + ( 2 x + 5 ) d d x ( x - 1 ) = ( x - 1 ) ( 2 ) + ( 2 x + 5 ) = 2 x - 2 + 2 x + 5 = 4 x + 3 (ii) d d x 1 ( 2 x + 1 ) = d d x ( 2 x + 1 ) - 1 = - 2 ( 2 x + 1 ) - 2 = - 2 ( 2 x + 1 ) 2 (iii) d d x 3 x + 4 4 x + 5 = ( 4 x + 5 ) d d x ( 3 x + 4 ) - ( 3 x + 4 ) d d x ( 4 x + 5 ) ( 4 x + 5 ) 2 = 12 x + 15 - 12 x - 16 ( 4 x + 5 ) 2 = - 1 4 x + 5 2 (iv) d d x x 2 x 3 + 1 = ( x 3 + 1 ) d d x ( x 2 ) - x 2 d d x ( x 3 + 1 ) ( x 3 + 1 ) 2 = 2 x 4 + 2 x - 3 x 4 ( x 3 + 1 ) 2 = 2 x - x 4 x 3 + 1 2

Obtain the differential coefficient of the following (i) x – 1 2 x + 5 (ii) 1 2 x + 1 (iii) 3 x + 4 4 x + 5 (iv) x 2 x 3 + 1

(i) d d x ( x - 1 ) ( 2 x + 5 ) Using differentiating by parts method: = ( x - 1 ) d d x ( 2 x + 5 ) + ( 2 x + 5 ) d d x ( x - 1 ) = ( x - 1 ) ( 2 ) + ( 2 x + 5 ) = 2 x - 2 + 2 x + 5 = 4 x + 3 (ii) d d x 1 ( 2 x + 1 ) = d d x ( 2 x + 1 ) - 1 = - 2 ( 2 x + 1 ) - 2 = - 2 ( 2 x + 1 ) 2 (iii) d d x 3 x + 4 4 x + 5 = ( 4 x + 5 ) d d x ( 3 x + 4 ) - ( 3 x + 4 ) d d x ( 4 x + 5 ) ( 4 x + 5 ) 2 = 12 x + 15 - 12 x - 16 ( 4 x + 5 ) 2 = - 1 4 x + 5 2 (iv) d d x x 2 x 3 + 1 = ( x 3 + 1 ) d d x ( x 2 ) - x 2 d d x ( x 3 + 1 ) ( x 3 + 1 ) 2 = 2 x 4 + 2 x - 3 x 4 ( x 3 + 1 ) 2 = 2 x - x 4 x 3 + 1 2

Evaluate the following integrals: ∫ x 15 d x

We know that: ∫ x n d x = x n + 1 n + 1 + c So, ∫ x 15 d x = x 15 + 1 15 + 1 + c = x 16 16 + c

Evaluate the following integrals ∫ x – 3 2 d x .

It is known that, ∫ x n d x = x n + 1 n + 1 + C , where n is any number other than - 1 and C is a constant. Now, integrate the given expression with respect to x , ∫ x - 3 2 d x = x - 3 2 + 1 - 3 2 + 1 + C = x - 1 2 - 1 2 + C = - 2 x + C

Evaluate the following integrals: ∫ 3 x – 7 + x – 1 d x .

It is known: ∫ x n d x = x n + 1 n + 1 + c and ∫ x - 1 d x = ln x Given, ∫ 3 x – 7 + x – 1 d x = ∫ 3 x – 7 d x + ∫ x – 1 d x = 3 ∫ x – 7 d x + ∫ x – 1 d x = 3 x – 7 + 1 – 7 + 1 + ln x + c = – x – 6 2 + log e x + c

Evaluate the following integrals ∫ x + 1 x 2 d x

It is known that ∫ x n d x = x n + 1 n + 1 + C , where n is any number other than - 1 and C is a constant. Now, ∫ x + 1 x 2 d x = ∫ x 2 + 1 x 2 + 2 × x × 1 x d x = ∫ x + 1 x + 2 d x = ∫ x d x + ∫ 1 x d x + 2 ∫ d x = x 2 2 + ln x + 2 x + C

Evaluate the following integrals ∫ x + 1 x d x

It is known that ∫ x n d x = x n + 1 n + 1 + C , where n is any number other than - 1 and C is a constant. And ∫ 1 x d x = log e x Now, integrate the given expression with respect to x . ∫ x + 1 x d x = ∫ x d x + ∫ 1 x d x = ∫ x 1 + 1 1 + 1 + log e x + C = x 2 2 + log e x + C