Question

\(\displaystyle \int \sin^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx\) is equal to

• A. \((x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^2+8x+13}{9}\right)+c\)
• B. \(\frac{3}{2}\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{2}\log\left(\frac{4x^2+8x+13}{9}\right)+c\)
• C. \((x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{2}\log(4x^2+8x+13)+c\)
• D. \(\frac{3}{2}(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log(4x^2+8x+13)+c\)

Hint:

Use identity \(\sin^{-1}\left(\frac{u}{\sqrt{u^2+a^2}}\right)=\tan^{-1}\left(\frac{u}{a}\right)\). Then integrate \(\tan^{-1}\) using integration by parts formula.

Answer 1

The Correct Option is A

Step 1: Simplify inside the inverse sine.
\[ \frac{2x+2}{\sqrt{4x^2+8x+13}} = \frac{2(x+1)}{\sqrt{4(x+1)^2+9}} \] Step 2: Use standard identity.
We know:
\[ \sin^{-1}\left(\frac{u}{\sqrt{u^2+a^2}}\right)=\tan^{-1}\left(\frac{u}{a}\right) \] Here:
\[ u=2(x+1),\quad a=3 \] So:
\[ \sin^{-1}\left(\frac{2(x+1)}{\sqrt{4(x+1)^2+9}}\right)=\tan^{-1}\left(\frac{2(x+1)}{3}\right) \] Hence integral becomes:
\[ \int \tan^{-1}\left(\frac{2x+2}{3}\right)dx \] Step 3: Use formula for \(\int \tan^{-1}(ax+b)\,dx\).
Standard result:
\[ \int \tan^{-1}(t)\,dx = x\tan^{-1}(t)-\frac{1}{2a}\ln(1+t^2)+C \] Here \(t=\frac{2x+2}{3}\), so \(dt/dx=2/3\).
Step 4: Apply final result.
We get:
\[ (x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^2+8x+13}{9}\right)+c \] Final Answer: \[ \boxed{(x+1)\tan^{-1}\left(\frac{2x+2}{3}\right)-\frac{3}{4}\log\left(\frac{4x^2+8x+13}{9}\right)+c} \]