${\displaystyle {\begin{aligned}\int n^{c}=&{\frac {1}{(c+1)}}n^{c+1}+{\frac {1}{2}}n^{c}+{\frac {c}{2}}An^{(c-1)}+{\frac {c\cdot (c-1)\cdot (c-2)}{2\cdot 3\cdot 4}}Bn^{(c-3)}\\&{+{\frac {c\cdot (c-1)\cdot (c-2)\cdot (c-3)\cdot (c-4)}{2\cdot 3\cdot 4\cdot 5\cdot 6}}Cn^{(c-5)}}\\&{+{\frac {c\cdot (c-1)\cdot (c-2)\cdot (c-3)\cdot (c-4)\cdot (c-5)\cdot (c-6)}{2\cdot 3\cdot 4\cdot 5\cdot 6\cdot 7\cdot 8}}Dn^{(c-7)}\cdots }\end{aligned}}}$

In your article, replace the image with:

<math>\begin{align}
\int n^{c} = &\frac{1}{(c+1)} n^{c+1}+\frac{1}{2} n^{c}+\frac{c}{2} A n^{(c-1)}+\frac{c \cdot (c-1) \cdot (c-2)}{2 \cdot 3 \cdot 4} B n^{(c-3)} \\
&{+\frac{c \cdot (c-1) \cdot (c-2) \cdot (c-3) \cdot (c-4)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6} C n^{(c-5)} } \\
&{+\frac{c \cdot (c-1) \cdot (c-2) \cdot (c-3) \cdot (c-4) \cdot (c-5) \cdot (c-6)}{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 \cdot 8} D n^{(c-7)} \cdots }
\end{align}</math>