Answer
$\frac{4^{1+\sqrt{2}}-1}{3(1+\sqrt{2})}$
Work Step by Step
$\int_0^1 (3x+1)^\sqrt{2}dx$
Let $u=3x+1$. Then $du=3dx$, and $dx=\frac{du}{3}$. The bounds are $3*0+1=1$ and $3*1+1=4$.
$=\int_{3*0+1}^{3*1+1}u^\sqrt{2}*\frac{du}{3}$
$=\frac{1}{3}\int_1^4 u^\sqrt{2}du$
$=\frac{1}{3}(\frac{u^{1+\sqrt{2}}}{1+\sqrt{2}})|_1^4$
$=\frac{1}{3}(\frac{4^{1+\sqrt{2}}}{1+\sqrt{2}}-\frac{1^{1+\sqrt{2}}}{1+\sqrt{2}})$
$=\frac{1}{3}(\frac{4^{1+\sqrt{2}}-1}{1+\sqrt{2}})$
$=\frac{4^{1+\sqrt{2}}-1}{3(1+\sqrt{2})}$
