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Integrals

7.5. Integration by Partial Fraction.

Q.1. Evaluate : ∫dx/(x³ + x² + x + 1)

Solution :

Let I = ∫dx/(x³ + x² + x + 1).
Here, x³ + x² + x + 1 = x² (x + 1) + (x + 1) = (x + 1)(x² + 1).
Therefore, 1/(x³ + x² + x + 1)
= 1/[(x + 1)(x² + 1)]
≡ A/(x + 1) + (Bx + C)/(x² + 1).
Therefore, 1 ≡ A(x² + 1) + (Bx + C)(x +1).
Putting x = 0, 1, –1, we get
1 = A + C
1 = 2A + 2B + 2C
1 = 2A.
Thus, A = 1/2, B = –1/2, C = 1/2.
Therefore, 1/(x3+ x² + x + 1) = (1/2)[1/(x+ 1)] + {(1/2)(– x + 1)}/(x2 + 1).
Thus I = ∫dx/(x³ + x² + x + 1)
= 1/2 ∫dx/(x + 1) – 1/2 ∫[x/(x² + 1)]dx + 1/2 ∫dx/(x² + 1)
=1/2 log | x + 1| – 1/4 log |x² + 1| + 1/2 tan ^-1 x + c [Ans.]

Q.2. Evaluate : ∫[cos x/{(1 – sin x)(2 – sin x)}] dx.

Solution :

We have, I = ∫[cos x/{(1 – sin x)(2 – sin x)}] dx
Put sin x = t => cos x dx = dt
Therefore, I = ∫dt/[(1 – t)(2 – t)]
= ∫[1/(1 – t) – 1/(2 – t) ] dt [Resolving into partial fraction]
= – log |1 – t| + log|2 – t| + c
= log|(2 – sin x)/(1 – sin x)| + c. [Ans.]

Q.3. Evaluate : ∫[(3x – 2)/{(x + 1)2(x + 3)}] dx.

Solution :

We have, (3x – 2)/{(x + 1)2(x + 3)} ≡ A/(x + 1) + B/(x + 1)2 + C/(x + 3) ------ (1)
Or, 3x – 2 ≡ A(x + 1)(x + 3) +B(x + 3) +C(x + 1)2
Putting x = –1, –3, 0 we get
– 5 = 2B => B = – 5/2
– 11 = 4C => C – 11/4
– 2 = 3A + 3B + C
– 2 = 3A – 15/2 – 11/2
– 2 = 3A – 41/4
=> 3A – 2 + 41/4 = 33/4
=> A = 11/4.
Substituting for A, B, C in (1) we get
(3x – 2)/{((x + 1)2(x + 3)} = 11/4. 1/(x + 1) – 5/2. 1/(x + 1)² – 11/4.1/(x + 3)
Integrating we get
∫[(3x – 2)/{(x + 1)2(x + 3)}]dx
= 11/4∫dx/(x + 1) – 5/2∫dx/(x + 1)² – 11/4∫dx/(x + 3)
= 11/4 log|(x + 1)/(x + 3)| + 5/2. 1/(x + 1) + c. [Ans.]

Paper By Mr. M.P.Keshari
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Ph No. : 09434150289