Question

If \[\int\frac{2x+3}{(x-1)(x^{2}+1)}\mathrm{d}x\] = \[=\log_{e}\left\{(x-1)^{\frac{5}{2}}\left(x^{2}+1\right)^{a}\right\}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\] where A is an arbitrary constant, then the value of a is

विकल्प

• \[\frac{5}{4}\]

• \[-\frac{5}{4}\]

• \[-\frac{5}{3}\]

• \[-\frac{5}{6}\]

Answer 1

\[-\frac{5}{4}\]

Explanation:

\[\mathrm{Let~I}=\int\frac{2x+3}{\left(x-1\right)\left(x^2+1\right)}\mathrm{d}x\]

\[\frac{2x+3}{\left(x-1\right)\left(x^2+1\right)}=\frac{\mathrm{A}}{x-1}+\frac{\mathrm{B}x+\mathrm{C}}{x^2+1}\]

∴ 2x + 3 = A (x² + 1) + (Bx + C) (x − 1)
= (A + B) x² + (C − B) x + (A − C)

∴ A + B = 0
C − B = 2
A − C = 3

On solving, we get

\[\mathrm{A}=\frac{5}{2}\], \[\mathrm{B}=\frac{-5}{2}\], \[\mathbf{C}=\frac{-1}{2}\]

\[\therefore\quad\mathrm{I}=\int\frac{2}{x-1}\mathrm{d}x+\int\frac{\frac{-5}{2}x-\frac{1}{2}}{x^2+1}\mathrm{d}x\]

\[=\frac{5}{2}\int\frac{1}{x-1}\mathrm{d}x-\frac{5}{2}\int\frac{x}{x^{2}+1}\mathrm{d}x-\frac{1}{2}\int\frac{1}{x^{2}+1}\mathrm{d}x\]

\[=\frac{5}{2}\log\left(x-1\right)-\frac{5}{4}\log\left(x^{2}+1\right)-\frac{1}{2}\tan^{-1}x+\mathrm{A}\]

\[=\log\left(x-1\right)^{\frac{5}{2}}+\log\left(x^{2}+1\right)^{\frac{-5}{4}}-\frac{1}{2}\tan^{-1}x+\mathrm{A}\]

\[=\log\left[\left(x-1\right)^{\frac{5}{2}}\left(x^2+1\right)^{\frac{-5}{4}}\right]-\frac{1}{2}\tan^{-1}x+\mathrm{A}\]