Question

Evaluate: \[ \int \frac{dx}{(x+1)\sqrt{x^2+1}} \]

• A. \( \frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{1+x}{1-x}\right) + c \)
• B. \( \frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{1-x}{1+x}\right) + c \)
• C. \( -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{1-x}{1+x}\right) + c \)
• D. \( -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{1+x}{1-x}\right) + c \)

Hint:

To solve integrals of the form \( \int \frac{dx}{(Ax+B)\sqrt{Cx^2+Dx+E}} \), consider the substitution \( Ax+B = \frac{1}{t} \). In cases where the radical involves \( \sqrt{x^2 \pm a^2} \) or \( \sqrt{a^2-x^2} \) and there are linear terms like \( (x+1) \) or \( (x-1) \) outside the radical, a substitution of the form \( x = \frac{1-y}{1+y} \) or \( x = \frac{1+y}{1-y} \) can often simplify the integral to a standard form, frequently leading to inverse hyperbolic functions.

Answer 1

The Correct Option is C

Step 1: Choose a suitable substitution.
The presence of terms like \((x+1)\) and \(\sqrt{x^2+1}\) suggests a substitution that simplifies both. A common effective substitution for such forms is \( x = \frac{1-y}{1+y} \).
Step 2: Differentiate the substitution and express terms in \(y\).
From \( x = \frac{1-y}{1+y} \):
Differentiate \(x\) with respect to \(y\):
\[ dx = \frac{d}{dy}\left(\frac{1-y}{1+y}\right) dy = \frac{-(1+y) - (1-y)}{(1+y)^2} dy = \frac{-1-y-1+y}{(1+y)^2} dy = \frac{-2}{(1+y)^2} dy \] Now, express the terms in the denominator in terms of \(y\):
\[ x+1 = \frac{1-y}{1+y} + 1 = \frac{1-y+1+y}{1+y} = \frac{2}{1+y} \] \[ x^2+1 = \left(\frac{1-y}{1+y}\right)^2 + 1 = \frac{(1-y)^2 + (1+y)^2}{(1+y)^2} = \frac{(1-2y+y^2) + (1+2y+y^2)}{(1+y)^2} = \frac{2+2y^2}{(1+y)^2} = \frac{2(1+y^2)}{(1+y)^2} \] So, \[ \sqrt{x^2+1} = \sqrt{\frac{2(1+y^2)}{(1+y)^2}} = \frac{\sqrt{2}\sqrt{1+y^2}}{|1+y|} \] Assuming \(1+y > 0\), we have \( \sqrt{x^2+1} = \frac{\sqrt{2}\sqrt{1+y^2}}{1+y} \).
Step 3: Substitute into the integral and simplify.
Substitute \(dx\), \((x+1)\), and \(\sqrt{x^2+1}\) into the integral: \[ \int \frac{\frac{-2}{(1+y)^2} dy}{\left(\frac{2}{1+y}\right)\left(\frac{\sqrt{2}\sqrt{1+y^2}}{1+y}\right)} \] Simplify the denominator: \[ \left(\frac{2}{1+y}\right)\left(\frac{\sqrt{2}\sqrt{1+y^2}}{1+y}\right) = \frac{2\sqrt{2}\sqrt{1+y^2}}{(1+y)^2} \] Now the integral becomes: \[ \int \frac{-2}{2\sqrt{2}\sqrt{1+y^2}} dy = -\frac{1}{\sqrt{2}} \int \frac{dy}{\sqrt{1+y^2}} \] Step 4: Evaluate the simplified integral and express in terms of \(x\).
The integral \( \int \frac{dy}{\sqrt{1+y^2}} \) is a standard integral: \[ \int \frac{dy}{\sqrt{1+y^2}} = \sinh^{-1}(y) + C' \] So the integral becomes: \[ -\frac{1}{\sqrt{2}} \sinh^{-1}(y) + C \] Finally, express \(y\) in terms of \(x\). From the substitution \( x = \frac{1-y}{1+y} \):
\[ x(1+y) = 1-y \] \[ x+xy = 1-y \] \[ xy+y = 1-x \] \[ y(x+1) = 1-x \] \[ y = \frac{1-x}{1+x} \] Substitute this back into the result: \[ -\frac{1}{\sqrt{2}} \sinh^{-1}\left(\frac{1-x}{1+x}\right) + C \]