MathematicsMedium210×since 2002Q3444Let In(x)=∫0x1(t2+5)ndt,n=1,2,3,….I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .In(x)=∫0x(t2+5)n1dt,n=1,2,3,…. Then :A50I6−9I5=xI5′50 I_{6}-9 I_{5}=x I_{5}^{\prime}50I6−9I5=xI5′B50I6−11I5=xI5′50 I_{6}-11 I_{5}=x I_{5}^{\prime}50I6−11I5=xI5′C50I6−9I5=I5′50 I_{6}-9 I_{5}=I_{5}^{\prime}50I6−9I5=I5′D50I6−11I5=I5′50 I_{6}-11 I_{5}=I_{5}^{\prime}50I6−11I5=I5′Check answerSkip