<s>[INST] Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
 [/INST] Would you not just introduce substitution $u=\sqrt{3x+1}$, $x=\frac{u^2-1}{3}$, $dx=\frac23 udu$ so:
$$\int (2x+3) \sqrt {3x+1} dx=\int \left(\frac23 (u^2-1)+3\right)u\cdot \frac23 udu$$
which is an integral of a polynomial in $u$ - easy to solve.
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