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Question Number 131527 by Sudip last updated on 05/Feb/21
Answered by Dwaipayan Shikari last updated on 05/Feb/21
27∫tt+227dt+57∫1t+227dt=27t−4449log(t+227)+57log(t+227)
Answered by mr W last updated on 05/Feb/21
∫2t+57t+22dt=17∫2×7t+2×22−97t+22dt=17∫(2−97t+22)dt=17(2t−∫97t+22dt)=17[2t−97∫17t+22d(7t+22)]=17[2t−97ln(7t+22)]+C=27t−949ln(7t+22)+C
Answered by Bird last updated on 06/Feb/21
I=∫2t+57t+22dtwedothech.7t+22=x⇒t=x−227⇒I=∫1x(2x−447+5)dx7=149∫1x(2x−44+35)dx=149∫2x−9xdx=249x−949ln∣x∣+C=249(7t+22)−949ln∣7t+22∣+C
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