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determine integral: S t^3/2t^4 -5 dt
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S t^3/2t^4 -5 dt = S t^3/2t^4 + S -5 dt = 1/2 S t^3/t^4 - 5 t = 1/2 S 1/t - 5t = 1/2 * ln t - 5 t + C
\[\int\limits \frac{t^3}{2t^4-5}dt\] Let u=2t^4-5 then du=8t^3 dt So:(1/8)du=t^3dt So your integral becomes: \[\frac{1}{8} \int\limits \frac{du}{u}=\frac{1}{8}\ln|u|+C=\frac{1}{8}\ln|2t^4-5|+C\] Note: If the "-5" is NOT in the denominator of the fraction. Then Ana2's method is correct.
yeah it is in the denominator of the fraction
Then my method would be the appropriate one :)
ok how do u know to do 1/8?
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Because you have 8t^3 but in the numerator you only have t^3. In other words, its off by a factor of 8. So you need to divide by 8.
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