CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Integrals

7.1. Integration as an Inverse Process of Differentiation.

Q.1. Evaluate : ∫[(x² + 1)/(x + 1)²]dx.

Solution :

We have, ∫[(x² + 1)/(x + 1)²]dx = ∫[{(x + 1)² – 2x }]dx
= ∫[{(x + 1)² – 2x(x + 1) + 2}/(x + 1)²]dx
= ∫[1 – 2/(x + 1) + 2/(x + 1)²]dx
= x – 2log| x + 1| –2/(x + 1) + c. [Ans.]

Q.2. Evaluate : ∫[(x + 3)/(x² + 4x + 3)] dx.

Solution :

We have, I = ∫[(x + 3)/(x² + 4x + 3)] dx
= ∫[(x +3)/{(x + 3)(x + 1)}] dx
= ∫[1/(x + 1)] dx
= log|x + 1| + c. [Ans.]

7.2. Integration by Substitution.

Q.1. Evaluate : ∫[x²/(1 + x³)] dx.

Solution :

We have, ∫[x²/(1 + x³)]dx = I (say)
Put 1 + x³ = t , then 3x²dx = dt
Therefore, I = 1/3∫dt/t
= 1/3 log t + c
= 1/3 log(1 + x3) + c. [Ans.]

Q.2. Evaluate : ∫[{2x. tan –1(x²)}/(1 + x⁴)] dx.

Solution :

Let I = ∫[{2x. tan –1(x²)}/(1 + x⁴)] dx.
Put tan –1(x²) = t , then [1/{1 + (x²)²}].2x dx = dt
Therefore, I = ∫t dt
= t²/2 + c
= 1/2 {tan –1(x²)}² + c. [Ans.]

Q.3. Evaluate : ∫[cos x/√(sin² x – 2sin x – 3)] dx.

Solution :

Let I = ∫[cos x/√(sin² x – 2 sin x – 3)] dx
Put sin x = t => cos x dx = dt
Therefore, I = ∫[dt/√(t² – 2t – 3)] dt
= ∫[dt/√{(t – 1)2 – 2²}]
= log|(t – 1) + √{(t – 1)2 – 2²}| + c
= log|(sin x – 1) + √(sin² x – 2 sin x – 3)| + c. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289