More practice questions

  1. Q5. Find ∫dx/√((2 − x)2 + 1) (NCERT Exercise 7.4, Q3, page 251).

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    Put t = 2 − x, so dx = −dt. The integral is −∫dt/√(t2 + 1) = −log |2 − x + √(x2 − 4x + 5)| + C, since (2 − x)2 + 1 = x2 − 4x + 5. (NCERT writes it as the log of the reciprocal, which is the same.)

  2. Q6. Find ∫3x/(1 + 2x4) dx (NCERT Exercise 7.4, Q5, page 251).

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    Put t = √2 x2, so dt = 2√2 x dx and 2x4 = t2. The integral is (3/(2√2))∫dt/(1 + t2) = (3/(2√2)) tan (√2 x2) + C.

  3. Q7. Find ∫x2/(1 − x6) dx (NCERT Exercise 7.4, Q6, page 251).

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    Put t = x3: (1/3)∫dt/(1 − t2) = (1/3) × (1/2) log |(1 + t)/(1 − t)|, that is (1/6) log |(1 + x3)/(1 − x3)| + C.

  4. Q8. Find ∫(x − 1)/√(x2 − 1) dx (NCERT Exercise 7.4, Q7, page 251).

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    Split the top: ∫x/√(x2 − 1) dx = √(x2 − 1), and ∫dx/√(x2 − 1) = log |x + √(x2 − 1)|. The answer is √(x2 − 1) − log |x + √(x2 − 1)| + C.

  5. Q9. Find ∫x2/√(x6 + a6) dx (NCERT Exercise 7.4, Q8, page 251).

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    Put t = x3: (1/3)∫dt/√(t2 + (a3)2) = (1/3) log |x3 + √(x6 + a6)| + C.

  6. Q10. Find ∫dx/√((x − 1)(x − 2)) (NCERT Exercise 7.4, Q13, page 252).

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    (x − 1)(x − 2) = x2 − 3x + 2 = (x − 3/2)2 − (1/2)2. With t = x − 3/2, the formula for ∫dt/√(t2 − a2) gives log |x − 3/2 + √(x2 − 3x + 2)| + C.

  7. Q11. Find ∫(x + 2)/√(4x − x2) dx (NCERT Exercise 7.4, Q20, page 252).

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    x + 2 = A(4 − 2x) + B gives A = −½ and B = 4. The A part is −½ × 2√(4x − x2) = −√(4x − x2). For the B part, 4x − x2 = 4 − (x − 2)2, so it is 4 sin ((x − 2)/2). The answer is −√(4x − x2) + 4 sin ((x − 2)/2) + C.

  8. Q12. Find ∫(x + 3)/(x2 − 2x − 5) dx (NCERT Exercise 7.4, Q22, page 252).

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    x + 3 = A(2x − 2) + B gives A = ½ and B = 4. The A part is ½ log |x2 − 2x − 5|. For the B part, x2 − 2x − 5 = (x − 1)2 − (√6)2, and with u = x − 1 that is u2 − 6, so the B part is 4 × 1/(2√6) = 2/√6 times log |(u − √6)/(u + √6)|. The answer is the sum of the two parts, plus C.