Answer: (a) \(\displaystyle \int \frac{x}{\sqrt{1+x^2}}\,dx=\sqrt{1+x^2}+C\).
(b) \(\displaystyle y=x^2\sinh^{-1}x-x\sqrt{1+x^2}+x\big(1+\sqrt2-\ln(1+\sqrt2)\big)\).
(a) Let \(u=1+x^2\). Then \(du=2x\,dx\), so \(x\,dx=\tfrac12 du\).
Hence
\(\displaystyle \int \frac{x}{\sqrt{1+x^2}}\,dx=\frac12\int u^{-1/2}\,du=\sqrt{u}+C=\sqrt{1+x^2}+C\).
(b) Start with
\(\displaystyle x\frac{dy}{dx}-y=x^2\sinh^{-1}x\).
Divide through by \(x\):
\(\displaystyle \frac{dy}{dx}-\frac{y}{x}=x\sinh^{-1}x\).
This is linear, with integrating factor
\(\displaystyle \mu(x)=e^{\int -1/x\,dx}=e^{-\ln x}=x^{-1}\).
Multiplying through by \(x^{-1}\) gives
\(\displaystyle \frac{d}{dx}\big(x^{-1}y\big)=\sinh^{-1}x\).
Integrate:
\(\displaystyle x^{-1}y=\int \sinh^{-1}x\,dx + C\).
Using integration by parts,
\(\displaystyle \int \sinh^{-1}x\,dx=x\sinh^{-1}x-\int \frac{x}{\sqrt{1+x^2}}\,dx\).
From part (a), \(\displaystyle \int \frac{x}{\sqrt{1+x^2}}\,dx=\sqrt{1+x^2}\), so
\(\displaystyle x^{-1}y=x\sinh^{-1}x-\sqrt{1+x^2}+C\).
Hence
\(\displaystyle y=x^2\sinh^{-1}x-x\sqrt{1+x^2}+Cx\).
Now use \(y=1\) when \(x=1\):
\(\displaystyle 1=\sinh^{-1}1-\sqrt2+C\).
Since \(\sinh^{-1}1=\ln(1+\sqrt2)\),
\(\displaystyle C=1+\sqrt2-\ln(1+\sqrt2)\).
Therefore
\(\displaystyle y=x^2\sinh^{-1}x-x\sqrt{1+x^2}+x\big(1+\sqrt2-\ln(1+\sqrt2)\big)\).
