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

Integrals

Q.4. Evaluate : ∫[(2x + 1)/(2x² + 4x – 3)] dx.

Solution :

We have, I = ∫[(2x + 1)/(2x² + 4x – 3)] dx (say).
Let 2x + 1 = Ad/dx(2x² + 4x – 3) + B = A(4x + 4) + B => A = 1/2, B = –1.
Therefore, I = ∫[{1/2(4x + 4) – 1 }/(2x² + 4x – 3)] dx + c
= 1/2∫[(4x + 4)/(2x² + 4x – 3)] dx – ∫dx/(2x² + 4x – 3)
= 1/2 I1 – I2 + c ---------------------------- (1)
Where I1 = ∫[(4x + 4)/(2x² + 4x – 3)] dx
= ∫dt/t [Put 2x² + 4x – 3 = t, then (4x + 4)dx = dt]
= log |t| = log|2x² + 4x – 3| ----------------------- (2)
and I2 = ∫dx/(2x² + 4x – 3)
= 1/2∫dx/(x² + 2x – 3/2)
= 1/2∫dx/[(x + 1)² – {√(5/2)}2]
= (1/2).1/[2√(5/2)] log|{x + 1 – √(5/2)}/{x + 1 + √(5/2)}|
= 1/(2√10). log|{x + 1 – √(5/2)}/{x + 1 + √(5/2)| ---------------- (3)
From (1), (2) and (3), we get
I = 1/2 log|2x² + 4x –3| –1/(2√10)log|{x + 1 – √(5/2)}/{x + 1 + √(5/2)|
+ c. [Ans.]

Q.5. Evaluate : ∫[(x + 3)/(x² – 2x – 5)] dx.

Solution :

Let I = ∫[(x + 3)/(x² – 2x – 5)] dx.
And let x + 3 = A + B d/dx(x2 – 2x – 5) = A + B (2x – 2)
Equating the coefficients of like terms, we get
2B = 1 => B = 1/2.
A – 2B = 3 => A – 2(1/2) = 3 => A = 3 + 1 = 4.
Therefore, I = ∫[{4 + 1/2(2x – 2)}/(x² – 2x – 5)] dx
= 4∫[1/(x² – 2x – 5)] dx + 1/2 ∫[(2x – 2)/(x² – 2x – 5)] dx
= I1 + I2.
I1 = 4∫[1/{(x – 1)² – 1 – 5}] dx
= 4 ∫[1/{(x – 1)² – (√6)2}] dx
= 4.(1/2√2)log|(x – 1 – √6)/(x – 1 + √6)|
= √2log|(x – 1 – √6)/(x – 1 + √6)|
I2 = 1/2 ∫[(2x – 2)/(x² – 2x – 5)] dx
[Put x² – 2x – 5 = t => (2x – 2)dx = dt]
= 1/2 ∫dt/t = 1/2 log|t| = 1/2 |x² – 2x – 5|
Therefore, I = I1 + I2 = √2log|(x – 1 – √6)/(x – 1 + √6| + 1/2 |x² – 2x – 5| + c. [Ans.]

Q.6. Evaluate : ∫[x²/(x⁴ + x² + 1)]dx.

Solution :

Let I = ∫[x²/(x4+ x² + 1)]dx
= 1/2∫[2x²/(x⁴ + x² + 1)]dx
= 1/2∫[(x² + 1 + x² – 1)/(x4 + x² + 1)]dx
= 1/2∫[(x² + 1)/(x⁴ + x² + 1)]dx + 1/2∫[(x² – 1)/(x⁴ + x² + 1)]dx
= 1/2 I1 + 1/2 I2.
Let us consider I1 = ∫[(x² + 1)/(x⁴ + x² + 1)]dx
= ∫[(1 + 1/x²)/(x² + 1 + 1/x²)]dx [Dividing Nr and Dr by x²]
= ∫[(1 + 1/x²)/{(x – 1/x)² + 2 + 1}]dx
= ∫[(1 + 1/x²)/{(x – 1/x)² + (√3)²}]dx
Put x – 1/x = t => (1 + 1/x²)dx = dt
I1 = ∫dt/[t² + (√3)²]
= 1/2√3tan –1(t/√3)
= 1/2√3tan –1{(x – 1/x)/√3}
= 1/2√3tan –1{(x² – 1)/x√3}
And I2 = ∫[(x² – 1)/(x⁴ + x² + 1)]dx
= ∫[(1 – 1/x²)/(x² + 1 + 1/x²)]dx [Dividing Nr and Dr by x²]
= ∫[(1 – 1/x²)/{(x + 1/x)² – 2 + 1 }]dx
= ∫[(1 – 1/x²)/{(x + 1/x) – 1}]dx
Put x + 1/x = t => (1 – 1/x²)dx = dt
I2 = ∫dt/(t² – 1)
= ∫dt/[t² – (1)²]
= 1/(2×1)log|(t – 1)/(t + 1)|
= 1/2log|(x + 1/x – 1)/(x + 1/x + 1)
= 1/2log|(x² + 1 – x)/(x² + 1 + x)|.
Therefore, I = (1/2)1/(2√3)tan –1{(x² – 1)/x√3}
+ (1/2)1/2log|(x² – x + 1)/(x² + x + 1)| + c. [Ans.]

Paper By Mr. M.P.Keshari
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