Evaluate the following integrals. Include absolute values only when needed. ∫ln2x+2ln⁡x−1xdx

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Evaluate the following integrals. Include absolute values only when needed.  ∫ln2x+2ln⁡x−1xdx

Pansdorfp6 · Accepted Answer

Step 1 given integral, ∫ln2x+2ln⁡x−1xdx we have to find the given integral. Step 2 here , ∫ln2x+2ln⁡x−1xdx substitute ln⁡x=t⇒1xdxdt=1⇒1xdx=dt ⇒∫ln2x+2ln⁡x−1xdx=∫(t2+2t−1)dt =∫t2dt+2∫tdt−∫1dt  =∫t2dt+2∫tdt−∫1dt  =t33+2t22−t+constant  =t33+t2−t+constant undo substitution t=ln⁡x ⇒∫ln2⁡x+2ln⁡x−1xdx=ln3⁡x3+ln2⁡x−ln⁡x+constant this is the required answer

Annie Gonzalez · Accepted Answer

ln2(x)+2ln⁡(x)−1xdx  Substitution u=ln⁡(x)⇒dudx=1x⇒dx=xdu, we use:  ln2(x)=u2  =∫(u2+2u−1)du  Lets

nick1337 · Accepted Answer

∫(ln⁡(x))2∗ln⁡(x)−1xdx Let us 1x∗dx=d(ln⁡(x)),t=ln⁡(x) put the expression 1 / x under the differential sign, i.e.: Then the original integral can be written as follows: ∫(t2+2∗t−1)∗dt This is a tabular integral: ∫t2+2t−1∗dt=t33+t2−t+C To write down the final answer, it remains to substitute log⁡(x) instead of t. ln⁡(x)33+ln⁡(x)2−ln⁡(x)+C

Evaluate the following integrals. Include absolute values only when needed. \int

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