Question

How do I find the integral
∫10(x−1)(x2+9)dxint10/((x-1)(x^2+9))dx ?

Answer 1

`\int(10)/((x-1)(x^2+9))\ dx=(*)\\\\(10)/((x-1)(x^2+9))=(A)/(x-1)+(Bx+C)/(x^2+9)=(A(x^2+9)+(Bx+C)(x-1))/((x-1)(x^2+9))\\\\=(Ax^2+9A+Bx^2-Bx+Cx-C)/((x-1)(x^2+9))=((A+B)x^2+(-B+C)x+(9A-C))/((x-1)(x^2+9))\\\Updownarrow\\10=(A+B)x^2+(-B+C)x+(9A-C)\\\Updownarrow\\A+B=0\ and\ -B+C=0\ and\ 9A-C=10\\A=-B\ and\ C=B\to9(-B)-B=10\to-10B=10\to B=-1\\A=-(-1)=1\ and\ C=-1`


`(*)=\int\left((1)/(x-1)+(-x-1)/(x^2+9)\right)\ dx=\int\left((1)/(x-1)-(x+1)/(x^2+9)\right)\ dx\\\\=\int(1)/(x-1)\ dx-\int(x+1)/(x^2+9)\ dx=\int(1)/(x-1)-\int(x)/(x^2+9)\ dx-\int(1)/(x^2+9)\ dx=(**)\\\\\#1\ \int(1)/(x-1)\ dx\Rightarrow\left|\begin{array}{ccc}x-1=t\\dx=dt\end{array}\right|\Rightarrow\int(1)/(t)\ dt=lnt+C_1=ln(x-1)+C_1`


`\#2\ \int(x)/(x^2+9)\ dx\Rightarrow \left|\begin{array}{ccc}x^2+9=u\\2x\ dx=du\\x\ dx=(1)/(2)\ du\end{array}\right|\Rightarrow\int\left((1)/(2)\cdot(1)/(u)\right)\ du=(1)/(2)\int(1)/(u)\ du\\\\\\=(1)/(2)ln(u)+C_2=(1)/(2)ln(x^2+9)+C_2`


`\#3\ \int(1)/(x^2+9)\ dx=\int(1)/(x^2+3^2)\ dx=(1)/(3)tan^(-1)\left((x)/(3)\right)+C_3\\\\therefore:\\\\\#1;\ \#2;\ \#3\Rightarrow(**)=ln(x-1)+C_1-(1)/(2)ln(x^2+9)+C_2-(1)/(3)tan^(-1)\left((x)/(3)\right)+C_3`


`\boxed{=ln(x-1)-(1)/(2)ln(x^2+9)-(1)/(3)tan^(-1)\left((x)/(3)\right)+C}`