Question

\(\begin{array}{l} I_n\left(x\right)=\int_0^x\frac{1}{\left(t^2+5\right)^n}dt, n=1, 2, 3,\cdots\end{array}\)

Then

• A. \(50I_6, - 9I_5 = xI'_5\)
• B. \(50I_6, - 11I_5 = xI'_5\)
• C. \(50I_6, - 9I_5 = I'_5\)
• D. \(50I_6, - 11I_5 = I'_5\)

Answer 1

The Correct Option is A

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

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

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

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

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

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

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

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