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Evaluate the following integral1.5 Tdx1.7* |
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Answer» / x/ x^4-a^4dx; x^2=t, 2x=dt; /1/2dt/t^2-(a^2)^2= =1/2 / dt/t^2-(a^2)^2; =1/2 / 1/2a^2ln|(-a^2)/ t+a^2)|+c =1/4a^2|x^2-a^2= x^2+ a^2|+c yes it is the correct answer / x/ x^4-a^4dx; x^2=t,2x=dt;. /1/2dt/t^2-(a^2)^2==1/2 / dt/t^2-(a^2)^2. =1/2 / 1/2a^2ln|(-a^2)/ t+a^2)|+c =1/4a^2-a^2=x^2+x |
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