Test: Indefinite Integration-Integration by Partial Fractions (23 Sep) - JEE MCQ

∫1/cos² x(1 - tanx)(2 - tanx) dx
= ∫(sec² x)/(2 - 3tanx + tan²x) dx
Put t = tanx 
dt = sec²x dx
∫dt/(2 - 3t + t²)dx
∫dt/(t² - 3t + 9/4) + (2 - 9/4)
= ∫dt/((t - 3/2)² - (½)
Using the formula ∫dx/(x² - a²), we get
= ½[½] log |[t - 3/2 - ½]/[t - 3/2 + ½]| + c
= log|(t - 2)/(t - 1)| + c
Taking - common, we get
= log|(2 - t)/(1 - t)| + c
log|(2 - tanx)/(1 - tanx)| + c

 ∫dx/x³(x^-2 -4).............(1)
=  ∫x^-3 dx/(x^-2 - 4)​
Let t = (x^-2 - 4)
dt = -2x^-3 dx
x^-3 = -dt/2
Put the value of x^-3 in eq(1)
= -½  ∫dt/t
= -½ log t + c
= -½ log(x^-2 - 4) + c
= -½ log(1-4x²)/x² + c
= ½ log(x²/(1 - 4x²)) + c

∫dx/x(xⁿ + 1)..............(1)
∫dx/x(xⁿ + 1) *(xⁿ - 1)/(xⁿ - 1)  
Put xⁿ = t
dt = nx^(n-1)dx
dt/n = x^(n-1)dx
Put the value of dt/n in eq(1)
= ∫(1/n)dt/t(t+1)
= 1/n ∫dt/(t+1)t
= 1/n{∫dt/t -  ∫dt/t+1}
= 1/n {ln t - lnt + 1} + c
= 1/n {ln |t/(t + 1)|} + c
= 1/n {ln |xⁿ/(xⁿ + 1)|} + c

Let I = ∫5x dx/[(x+1)(x2+9)]
Let 5x/[(x+1)(x2+9)] = A(x+1) + (Bx + C)/(x²+9)
⇒5x = (A+B)x² + (B+C)x + 9A + C
Comparing the coefficients of x2 on both sides, we get
A + B = 0    ...(1)
Comparing the coefficients of x on both sides, we get
B + C = 5    .....(2)
Comparing constants on both sides, we get
9A +C = 0    ....(3)
From (1), we get B = −A
From (3), we get C = −9A
now, from (2), we get 
−A − 9A = 5 
⇒A = −1/2
B = 1/2
C = 9/2
So, 5x/[(x+1)(x²+9)] = −1/2 × [1(x+1)] + 1/2 × (x + 9)/(x² + 9)
So, ∫5x dx/[(x+1)(x² + 9)] = −1/2∫dx/(x+1) + 1/2∫(x+9)/(x² +9) dx
=−1/2 log|x+1| + 1/2∫x dx/(x² +9) + 9/2∫dx (x² +9)
=−1/2 log|x+1| + 1/4∫2x dx/(x² +9) + 9/2∫dx/(x² +(3)²)
⇒−1/2 log |x+1| + 1/4log|x ² +9| + 9/2×1/3 tan⁻¹(x/3) + C
=−1/2 log |x+1| + 1/4log∣|x2+9| + 3/2 tan⁻¹(x/3) + C

∫dx/(x² - a⁴b⁴)
= ∫dx/(x² - a²b²)²
= ∫dx/(x² - a²b²)(x² + a²b²)
As by formula dx/(n - n)(n + n) = 1/2n log|(x - n)/(x + n)|
= 1/2n log|(x - a²b²)/(x + a²b²)|