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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.
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Question : 198 of 271
Marks: +1, -0
∫01xx2+1dx\int\limits_{0}^{1} \frac{x}{x^2+1}dx
Solution:  
Let I=∫01xx2+1dxI=\int\limits_{0}^{1} \frac{x}{x^2+1}dx
Put x2+1=tx^2 + 1 = t ⇒2xdx=dt\Rightarrow 2x dx = dt; When x=0,t=1x = 0, t = 1 and when x=1,t=2x = 1, t = 2
∴I=12∫01dtt=[12log⁡t]12\therefore I=\frac{1}{2}\int\limits_{0}^{1} \frac{dt}{t}=\left[\frac{1}{2}\log t\right]_{1}^{2}
=12[log⁡2−log⁡1]=12log⁡2=\frac{1}{2}\left[\log 2 - \log 1\right]=\frac{1}{2}\log 2
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