_{-1}¹ (1+x)/1+x² dx = ₀¹ (1+x)/1+x² dx + ₀¹ f(x) dx, then f(x) =

Correct Answer is : log⁡(1−x)1+x2(1-x)/1+x²1+x2log(1−x)

Correct Answer is : log⁡(1−x)1+x2(1-x)/1+x²1+x2log(1−x)

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TS EAMCET 4-Aug-2021 Shift 1 Question Paper

Section: Mathematics

Question : 79 of 160

Marks: +1, -0

\int\limits_{-1}^{1} \frac{\log(1+x)}{1+x^2}

dx

\int\limits_{0}^{1} \frac{\log(1+x)}{1+x^2}

dx

\int\limits_{0}^{1} f(x) dx

f(x)

Solution: 👈: Video Solution

\int\limits_{-1}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx = \int\limits_{0}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx + \int\limits_{0}^{1} f(x) dx

\Rightarrow \int\limits_{-1}^{0} \;\frac{\log(1+x)}{1+x^{2}} dx + \int\limits_{0}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx

\;\; = \int\limits_{0}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx + \int\limits_{0}^{1} f(x) dx

\Rightarrow \int\limits_{-1}^{0} \;\frac{\log(1+x)}{1+x^{2}} dx + \int\limits_{0}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx

\;\; = \int\limits_{0}^{1} \;\frac{\log(1+x)}{1+x^{2}} dx = \int\limits_{0}^{1} f(x) dx

\Rightarrow \;\; \int\limits_{-1}^{0} \;\frac{\log(1+x)}{1+x^{2}} dx = \int\limits_{0}^{1} f(x) dx

\;\text{ Let }\; I = \int\limits_{0}^{1} f(x) dx = \int\limits_{-1}^{0} \;\frac{\log(1+x)}{1+x^{2}} dx

\;\text{ Put }\; x = -y

\Rightarrow dx = -dy

\;\text{ For }\; x=-1,\; y=1,\; x=0,\; y=0

\therefore \;\; I = \int\limits_{1}^{0} \;\frac{\log(1-y)}{1+y^{2}}(-dy)

\;\;\; = \int\limits_{0}^{1} \;\frac{\log(1-y)}{1+y^{2}} dy

By change of variable property,

I \;\; = \int\limits_{0}^{1} \;\frac{\log(1-x)}{1+x^{2}} dx

\;\text{ or }\; \int\limits_{0}^{1} f(x) dx \;\; = \int\limits_{0}^{1} \;\frac{\log(1-x)}{1+x^{2}} dx

f(x) \;\; = \frac{\log(1-x)}{1+x^{2}}

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