If \;1-x^7/x(1+x^7) dx = a |x|+b x^7+1|+c then

Correct Answer is : a=1a = 1a=1 and b= −27b = \;-2/7b=−72

Correct Answer is : a=1a = 1a=1 and b= −27b = \;-2/7b=−72

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CGPET 2022 Solved Paper

Section: Mathematics

Question : 141 of 150

Marks: +1, -0

\int\;\frac{1-x^7}{x(1+x^7)} dx = a \ln|x|+b \ln x^7+1|+c

Solution:

Let

I = \int\frac{1-x^7}{x(1+x^7)} dx

= \int\frac{2-(1+x^7)}{x(1+x^7)} dx

= 2 \int\frac{1}{x(1+x^7)} dx - \int\frac{1}{x} dx

= 2 \int\frac{x^6}{x^7(1+x^7)} dx - \ln x + c_1

Again, let

I_1 = \int\frac{x^6}{x^7(1+x^7)} dx

Put

x^7 = t

and

7x^6 dx = dt

\therefore I_1 = \frac{1}{7} \int\frac{dt}{t(1+t)}

= \frac{1}{7} \int \left(\frac{1}{t} - \frac{1}{1+t}\right) dt

= \frac{1}{7} \ln t - \frac{1}{7} \ln(1+t) + c_2

= \frac{1}{7} \ln x^7 - \frac{1}{7} \ln(1+x^7) + c_2

= \ln x - \frac{1}{7} \ln(1+x^7) + c_2

Now,

I = 2I_1 - \ln x + c_1

= 2 \ln x - \frac{2}{7} \ln(1+x^7) - \ln x + c_1 + 2c_2

= \ln x - \frac{2}{7} \ln(1+x^7) + c,

where,

c = c_1 + 2c_2

\therefore a = 1, b = -\frac{2}{7}

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