Evaluate the indefinite integral as an infinite series. ∫ex−1xdx

Question

Evaluate the indefinite integral as an infinite series.  ∫ex−1xdx

Piosellisf · Accepted Answer

We know that the Maclaurin series for ex is:ex=∑n=0∞xnn!=1+∑n=1∞xnn!⇒ex−1=∑n=1∞xnn!Furthermore, x can be considered as a constant in regards to n.That&#039;s why we heve (when x≠0):ex−1x=1x⋅(ez−1)=1x⋅∑n=1∞xnn!=∑n=1∞xn−1n!Integrating is now easy:∫ex−1xdx=∫∑n=1∞xn−1n!=∑n=1∞1n!∫xn−1dx=∑n=1∞1n!xnn+C=∑n=1∞xnn⋅n!+C=∑n=0∞xn+1(n+1)⋅(n+1)!+C∫ex−1xdx∑n=0∞xn+1(n+1)⋅(n+1)!+C

hysgubwyri3 · Accepted Answer

We know that  ex=∑n=0∞xnn!  This can be written as  ex=x00!+∑n=1∞xnn!  ex=1+∑n=1∞xnn!  Subtract 1 from both sides  ex−1=∑n=1∞xnn!  Divide both sides by x  ex−1x=∑n=1∞xn−1n!  Integrate both sides with respect to x  ∫ex−1xdx=∑n=1∞∫xn−1n!dx  In right side, we will use power rule for integration  ∫ex−1xdx=C+∑n=1∞∫xnn⋅n!

nick1337 · Accepted Answer

∫e2−1xdx=(e2−1)ln⁡|x|+C Steps ∫e2−1xdx Take the constant out: ∫a⋅f(x)dx=a⋅∫f(x)dx =(e2−1)⋅∫1xdx ∫1xdx=ln⁡|x| =(e2−1)ln⁡|x| Add a constant to the solution =(e2−1)ln⁡|x|+C Plotting: (e2−1)ln⁡|x|+C assuming C=0

Evaluate the indefinite integral as an infinite series. \int e^{x}-1/xdx

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New Questions in Calculus 1