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Mathematics Question on Integration by Parts

In(x)=∫0x1(t2+5)ndt,n=1,2,3,⋯\begin{array}{l} I_n\left(x\right)=\int_0^x\frac{1}{\left(t^2+5\right)^n}dt, n=1, 2, 3,\cdots\end{array}

Then

A

50I6,−9I5=xI5′50I_6, - 9I_5 = xI'_5

B

50I6,−11I5=xI5′50I_6, - 11I_5 = xI'_5

C

50I6,−9I5=I5′50I_6, - 9I_5 = I'_5

D

50I6,−11I5=I5′50I_6, - 11I_5 = I'_5

Answer

50I6,−9I5=xI5′50I_6, - 9I_5 = xI'_5

Explanation

Solution

In(x)=∫0x1(t2+5)ndt\begin{array}{l} I_n\left(x\right)=\displaystyle\int\limits_0^x\frac{1}{\left(t^2+5\right)^n}dt\end{array}

=∫0x1(t2+5)n⏟I×I⏟IIdt\begin{array}{l} =\displaystyle\int\limits_0^x\frac{1}{\underset{I}{\underbrace{\left(t^2+5\right)^n}}}\times\underset{II}{\underbrace{I}}dt\end{array}

=t(t2+5)n∣0x−∫0x−2nt(t2+5)n+1×t dt\begin{array}{l} \left.=\frac{t}{\left(t^2+5\right)^n} \right|^x_0-\displaystyle\int\limits_0^x\frac{-2nt}{\left(t^2+5\right)^{n+1}}\times t\ dt\end{array}

=x(x2+5)n+∫0x2n(t2+5−5(t2+5)n+1)dt\begin{array}{l} =\frac{x}{\left(x^2+5\right)^n}+\displaystyle\int\limits_0^x2n\left(\frac{t^2+5-5}{\left(t^2+5\right)^{n+1}}\right)dt\end{array}

In(x)=x(x2+5)n+2n In(x)−10n In+1(x)\begin{array}{l} I_n\left(x\right)=\frac{x}{\left(x^2+5\right)^n}+2n\ I_n\left(x\right)-10n\ I_{n+1}\left(x\right) \end{array}

10n In+1(x)−(2n−1) In(x)=xI’n(x)\begin{array}{l} 10n\ I_{n+1}\left(x\right)-\left(2n-1\right)\ I_n\left(x\right)=xI’_n\left(x\right) \end{array}

For n=5\50I6(x)−9I5(x)=xI’5(x)\begin{array}{l} \text{For}~n=5\\\50I_6\left(x\right)-9I_5\left(x\right)=xI’_5\left(x\right)\end{array}

For n=5\50I6(x)−9I5(x)=xI’5(x)\begin{array}{l} \text{For}~n=5\\\50I_6\left(x\right)-9I_5\left(x\right)=xI’_5\left(x\right)\end{array}