How you integrate $\frac{1}{\sqrt{1+x^2}}$ using following substitution? $1+x^2=t$ $\Rightarrow$ $x=\sqrt{t-1} \Rightarrow dx = \frac{dt}{2\sqrt{t-1}}dt$… Now I'm stuck. I don't know how to proceed using substitution rule.
Integration – How to Integrate $\frac{1}{\sqrt{1+x^2}}$ Using Substitution
indefinite-integralsintegrationsubstitution
Related Solutions
$$ \int \frac{8}{16-e^{4x}} \, dx=\left| u=e^{2x} \atop du=2e^{2x}\,dx=2u\, dx \right| =\int\frac{8}{16-u^2} \cdot \frac{du}{2u} $$ (we assume that $u\in(-4,4)$) $$ =\left| u=4\sin t \atop du=4\cos t\,dt \right|= \int\frac{8}{16-16\sin^2t} \cdot \frac{4\cos t\,dt}{8\sin t} =\frac14 \int \frac{\cos t\,dt}{\cos^2t \sin t} $$ $$ =\frac12 \int \frac{dt}{2\sin t\cos t}= \frac12 \int \frac{dt}{\sin 2t}= \left| z= 2t \atop dz= 2\,dt \right|= $$ $$ =\frac14 \int \frac{dz}{\sin z}= \frac14\ln\left| \tan\left( \frac{z}2 \right) \right|+C =\frac14\ln\left| \tan t \right|+C $$ $$ =\frac14\ln\left| \tan\left( \arcsin\frac{u}4 \right) \right|+C =\frac14\ln\left| \tan\left( \arcsin\frac{e^{2x}}4 \right) \right|+C $$ Since $$ \tan^2(\arcsin x)=\frac1{\cot^2(\arcsin x)}= \frac1{\frac1{\sin^2(\arcsin x)}-1}= \frac1{\frac1{x^2}-1}=\frac{x^2}{1-x^2} $$ and (considering the sign) $$ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $$ the answer is $$ \frac14\ln\left| \frac{e^{2x}/4}{1-\sqrt{\frac{e^{4x}}{16}}} \right|+C =\frac{x}2-\frac14\ln\left| \sqrt{16-e^{4x}} \right|+C $$ $$ =\frac{x}2-\frac18\ln\left| 16-e^{4x} \right|+C. $$
One way to proceed is: \begin{align} \int x^2 \sqrt{x^2+1} \, dx &= \frac{1}{2} \int \sqrt{u} \sqrt{u-1} \, du \qquad (u = x^2 + 1) \\ &= - \frac{1}{16} \int \frac{y^8 - 2 y^4 + 1}{y^5} \, dy \qquad (y = \sqrt{u} - \sqrt{u-1}) \\ &= - \frac{y^4}{64} + \frac{1}{64 y^4} + \frac{\log(y)}{8} + C. \end{align}
Here, to change the integration variable from $u$ to $y$, one needs \begin{equation} \frac{dy}{du} = \frac{1}{2} \left( \frac{1}{\sqrt{u}} - \frac{1}{\sqrt{u-1}} \right) = - \frac{y}{2 \sqrt{u} \sqrt{u-1}}, \end{equation} and also the expression of $u$ in terms of $y$, which can be obtained by squaring the both sides of $\sqrt{u-1} = \sqrt{u}-y$ and then solving it for $\sqrt{u}$: \begin{equation} u = \frac{(y^2+1)^2}{4y^2}. \end{equation} So, after the variable transformation from $u$ to $y$, one gets a factor $u(u-1)$, which is rewritten as \begin{equation} u(u-1) = \frac{(y^2+1)^2(y^2-1)^2}{16 y^4}. \end{equation}
Now, after performing the integration, one needs the following substitutions \begin{equation} y = \sqrt{x^2 + 1} - x , \qquad y^4 - \frac{1}{y^4} = - 8 x (2 x^2 + 1) \sqrt{x^2 + 1}, \end{equation} to get the final result in $x$: \begin{equation} \int x^2 \sqrt{x^2+1} \, dx = \frac{1}{8} \left[ x (2x^2+1) \sqrt{x^2+1} + \log(\sqrt{x^2+1} - 1) \right] + C. \end{equation} Note that one can rewrite the log term as: \begin{equation} \log(\sqrt{x^2+1}-x) = - \log(\sqrt{x^2+1}+x) = - \sinh^{-1}(x). \end{equation}
Best Answer
By the substitution you suggested you get $$ \int \frac1{2\sqrt{t(t-1)}} \,dt= \int \frac1{\sqrt{4t^2-4t}} \,dt= \int \frac1{\sqrt{(2t-1)^2-1}} \,dt $$ Now the substitution $u=2t-1$ seems reasonable.
However your original integral can also be solved by $x=\sinh t$ and $dx=\cosh t\, dt$ which gives $$\int \frac{\cosh t}{\cosh t} \, dt = \int 1\, dt=t=\operatorname{arcsinh} x = \ln (x+\sqrt{x^2+1})+C,$$ since $\sqrt{1+x^2}=\sqrt{1+\sinh^2 t}=\cosh t$.
See hyperbolic functions and their inverses.
If you are familiar (=used to manipulate) with the hyperbolic functions then $x=a\sinh t$ is worth trying whenever you see the expression $\sqrt{a^2+x^2}$ in your integral ($a$ being an arbitrary constant).