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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.9
Evaluate the definite integrals in Exercises 176 to 195.
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Question : 194 of 271
Marks: +1, -0
∫026x+3x2+4dx\int\limits_{0}^{2} \frac{6x+3}{x^2+4} dx
Solution:  
Let I=∫026x+3x2+4dxI = \int\limits_{0}^{2} \frac{6x+3}{x^2+4} dx =∫026xx2+4dx+∫023x2+4dx= \int\limits_{0}^{2} \frac{6x}{x^2+4} dx + \int\limits_{0}^{2} \frac{3}{x^2+4} dx
=3∫022xx2+4dx+[3×12tan⁡−1x2]02= 3 \int\limits_{0}^{2} \frac{2x}{x^2+4} dx + \left[ 3 \times \frac{1}{2} \tan^{-1} \frac{x}{2} \right]_{0}^{2}
Let I1=3∫022xx2+4dxI_1 = 3 \int\limits_{0}^{2} \frac{2x}{x^2+4} dx
Put x2+4=t⇒2x dx=dtx^2 + 4 = t \Rightarrow 2x \, dx = dt
When x=0⇒t=4x = 0 \Rightarrow t = 4
When x=2⇒t=8x = 2 \Rightarrow t = 8
3∫48dtt=[3log⁡t]483 \int\limits_{4}^{8} \frac{dt}{t} = \left[ 3 \log t \right]_{4}^{8}
=3(log⁡8−log⁡4)=3log⁡2= 3(\log 8 - \log 4) = 3 \log 2
∴I=3log⁡2+32[tan⁡−11−tan⁡−10]\therefore I = 3 \log 2 + \frac{3}{2} \left[ \tan^{-1} 1 - \tan^{-1} 0 \right]
=3log⁡2+32×π4= 3 \log 2 + \frac{3}{2} \times \frac{\pi}{4}
=3log⁡2+3π8= 3 \log 2 + \frac{3\pi}{8}
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