Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
 A: Hint. One may find $a,b \in \mathbb{R}$ such that
$$
 (2x+3) \sqrt {3x+1}=\color{red}{a}\cdot(3x+1)^{3/2}+\color{red}{b}\cdot(3x+1)^{1/2}.
$$
A: Let us just split $2x+3 = \frac23 (3x+1)+\frac73$ and simplify to give us: $$I = \int (2x+3)\sqrt{3x+1}\, dx = \frac23 \int (3x+1)^{\frac32}\, dx+ \frac73 \int \sqrt {3x+1}\, dx$$ which can be easily solved.
A: $$\int  \left( 2x+3 \right) \sqrt { 3x+1 } dx=\\ 3x+1={ t }^{ 2 }\\ x=\frac { { t }^{ 2 }-1 }{ 3 } \\ dx=\frac { 2t }{ 3 } dt\\ \int { \left( \frac { 2{ t }^{ 2 }-2 }{ 3 } +3 \right)  } { t }\frac { 2t }{ 3 } dt=\frac { 2 }{ 9 } \int { \left( 2{ t }^{ 2 }+7 \right) { t }^{ 2 }dt } =\frac { 4 }{ 9 } \int { { t }^{ 4 }dt } +\frac { 14 }{ 9 } \int { { t }^{ 2 }dt } =\\ =\frac { 4 }{ 9 } \cdot \frac { { t }^{ 5 } }{ 5 } +\frac { 14 }{ 9 } \cdot \frac { { t }^{ 3 } }{ 3 } +C=\frac { 4 }{ 45 } { t }^{ 5 }+\frac { 14 }{ 27 } { t }^{ 3 }+C=\frac { 4 }{ 45 } { \left( 3x+1 \right)  }^{ \frac { 5 }{ 2 }  }+\frac { 14 }{ 27 } { \left( 3x+1 \right)  }^{ \frac { 3 }{ 2 }  }+C$$
A: By parts: $u=2x+3$ and $\mathrm dv=\sqrt{3x+1}$, then 
$$
\int(2x+3)\sqrt{3x+1}\;\mathrm dx=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{4}{9}\int(3x+1)^{3/2}\;\mathrm dx\\
=\frac{2}{9}(2x+3)(3x+1)^{3/2}-\frac{8}{135}(3x+1)^{5/2}+C
$$
A: 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.
