<s>[INST] Indefinite integration gives two different answers The question is :
\begin{array}{l}
\text { If } \int \frac{x+1}{\sqrt{2 x-1}} d x=\mathrm{f}(\mathrm{x}) \sqrt{2 x-1}+\mathrm{C}, \text { where } \mathrm{C} \text { is }\\
\text { constant of integration, then } \mathrm{f}(\mathrm{x}) \text { is equal to : }
\end{array}
Here is how we reach the answer :
\begin{array}{l}
\sqrt{2 x-1}=t \Rightarrow 2 x-1=t^{2} \Rightarrow 2 d x=2 t \\
. d t \\
\int \frac{x+1}{\sqrt{2 x-1}} d x=\int \frac{\frac{t^{2}+1}{2}+1}{t} t d t=\int \frac{t^{2}+3}{2} d t \\
=\frac{1}{2}\left(\frac{t^{3}}{3}+3 t\right)=\frac{t}{6}\left(t^{2}+9\right)+c \\
=\sqrt{2 x-1}\left(\frac{2 x-1+9}{6}\right)+c \\
=\sqrt{2 x-1}\left(\frac{x+4}{3}\right)+c \\
\Rightarrow f(x)=\frac{x+4}{3}
\end{array}
Here is how I solved it .
\begin{aligned}
& \int \frac{x+1}{\sqrt{2 x-1}} \\
\text { Let } 2 x-1 &=t \\
& \frac{d t}{d x}=2 \\
& x+1=\frac{t+3}{2} \\
&=\frac{1}{2} \int \frac{t+3}{2 \sqrt{t}} d t \\
&=\frac{1}{4} \int\left(\sqrt{t}+\frac{3}{\sqrt{t}}\right) d t \\
&=\frac{1}{4} \times \frac{2}{3} t^{3 / 2}+\frac{1}{4} \times \frac{2 \times 3}{1} \times t^{1 / 2} \\
=& \frac{1}{2}\left(\frac{1}{3} t^{3 / 2}+3 t^{1 / 2}\right) \\
&=t^{1 / 2}\left(\frac{1}{6} t^{3}+\frac{3}{2}\right) \\
&=\sqrt{2 x-1}\left(\frac{1}{6}(2 x-1)^{3}+\frac{3}{2}\right)
\end{aligned}
PS- It took me hour to write this . Pls tell if i solved it wrong . But Please answer.
 [/INST] $ \frac{1}{2}\left(\frac{1}{3} t^{3 / 2}+3 t^{1 / 2}\right) \\
=t^{1 / 2}\left(\frac{1}{6} t^{3}+\frac{3}{2}\right) $
This bit is wrong.
$t^{1/2} \times t^{3} \neq t^{3/2}.$
</s>