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Evaluation of Following Indefinite Integrals

\(\displaystyle (a)\;\; \int\frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2\cdot (x+1)}dx\)

\(\displaystyle (b)\;\; \int e^{m\cdot \sin^{-1}(x)}dx\)

(b)
First of all we apply the sub \( u=\sin^{-1} x \) to get: \( \displaystyle \int e^{mu}\cos u\,du \).
Now apply IBP to get:

$$\begin{aligned}
\int e^{mu}\cos u\, du &=\frac{e^{mu}\cos u}{m}+\int e^{mu}\sin u\, du \\
&= \frac{e^{mu}\cos u}{m}+\frac{e^{mu}\sin u}{m^2}-\frac{1}{m^2}\int e^{mu}\cos u\, du \\
\end{aligned}$$

Now:
$$\left ( \frac{1}{m^2}+1 \right )\int e^{mu}\cos u\, du=\frac{e^{mu}\sin u}{m^2}+\frac{e^{mu}\cos u}{m} \iff \int e^{mu}\cos u \, du=\frac{\frac{e^{mu}\sin u}{m^2}+\frac{e^{mu}\cos u}{m}}{\frac{1}{m^2}+1}$$

Substitute back for \(u=\sin^{-1} x \) to get:
$$\int e^{m\sin^{-1}x}\, dx=\frac{\left ( m\sqrt{1-x^2}+x \right )e^{m\sin^{-1}x}}{m^2+1}+c, \,\,\,\,c\in \mathbb{R}$$

I do not see a clear pattern for (a).

Nice Solution Apostolos J. Kos.

** My Try for \(\displaystyle (a)\;\; I = \int\frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2\cdot (x+1)}dx\)

Multiply both \(\bf{N_{r}}\) and \(\bf{D_{r}}\) by \(x+1\), We Get

\(\displaystyle I = \int\frac{(x^2-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)^2}dx = \int\frac{\left(1-\frac{1}{x^2}\right)\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}}{\left(x+\frac{1}{x}+2\right)}dx\)

So \(\displaystyle I = \int\frac{\left(1-\frac{1}{x^2}\right)\sqrt{\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)-3}}{\left(x+\frac{1}{x}+2\right)}dx\)

Now Substitute \(\left(x+\frac{1}{x}\right) = t\;,\) Then \(\left(1-\frac{1}{x^2}\right)dx = dt\), So \(\displaystyle I = \int\frac{\sqrt{t^2+2t-3}}{t+2}dt\)

Good job, Jacks... The last integral can be computed in latter by completing the square in the numinator , and then apply that sub. After some operations we'll get our desired result...