Test Index

NCERT Class XII Mathematics Chapter - - Solutions

Show Para  Hide Para 
Exercise - 7.9
Evaluate the definite integrals in Exercises 176 to 195.
© examsnet.com
Question : 189 of 271
Marks: +1, -0
∫012x+35x2+1dx\int\limits_{0}^{1}\frac{2x+3}{5x^2+1}dx
Solution:  
∫012x+35x2+1dx\int\limits_{0}^{1}\frac{2x+3}{5x^2+1}dx
=∫01(2x5x2+1+35x2+1)dx=\int\limits_{0}^{1}\left(\frac{2x}{5x^2+1}+\frac{3}{5x^2+1}\right)dx
=15∫0110x5x2+1dx+35∫01dxx2+(15)2=\frac{1}{5}\int\limits_{0}^{1}\frac{10x}{5x^2+1}dx+\frac{3}{5}\int\limits_{0}^{1}\frac{dx}{x^2+\left(\frac{1}{\sqrt{5}}\right)^2}
=15[log⁡(5x2+1)]01+35×115[tan⁡−1(x15)]01=\frac{1}{5}\left[\log(5x^2+1)\right]_{0}^{1}+\frac{3}{5}\times\frac{1}{\frac{1}{\sqrt{5}}}\left[\tan^{-1}\left(\frac{x}{\frac{1}{\sqrt{5}}}\right)\right]_{0}^{1}
=15(log⁡6−log⁡1)+35(tan⁡−15−0)=\frac{1}{5}(\log 6 - \log 1) + \frac{3}{\sqrt{5}}(\tan^{-1}\sqrt{5} - 0)
=15log⁡6+35tan⁡−15=\frac{1}{5} \log 6 + \frac{3}{\sqrt{5}} \tan^{-1}\sqrt{5}
© examsnet.com
Go to Question:
Prev Question
Next Question
Advertisements