Question

if$\int {\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$$=(x+1)ta{n}^{-1}\left(\frac{2x+2}{3}\right)$+$\lambda$$ln(4x^2+8x+13)$+$C$
then find the value of - 4$\lambda$

Answer 1

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Given

${\sin }^{-1}\left(\frac{2x+2}{\sqrt{4x^2+8x+13}}\right)$

${\sin }^{-1}\left(\frac{2x+2}{\sqrt{(2x+2{)}^2+(3{)}^2}}\right)$

${\tan }^{-1}\left(\frac{2x+2}{3}\right)$

$\int {\tan }^{-1}\left(\frac{2(x+1)}{3}\right)dx$

Put $t=1+x⟹dt=dx$

$\int {\tan }^{-1}\left(\frac{2t}{3}\right)$

Integrating by parts

$⟹{\tan }^{-1}\frac{2t}{3}\int dt-\int (\frac{d}{dt}{\tan }^{-1}\frac{2t}{3}\times \int dt)$

$⟹t{\tan }^{-1}\frac{2t}{3}-\int (\frac{1}{1+\frac{4t^2}{9}}\times \frac{2}{3}t)dt$

$⟹t{\tan }^{-1}\frac{2t}{3}-\frac{6}{8}\int \left(\frac{8tdt}{9+4t^2}\right)dt$

$⟹t{\tan }^{-1}\frac{2t}{3}-\frac{3}{4}\ln (9+4t^2)+c$

$⟹(1+x){\tan }^{-1}\frac{2x+2}{3}-\frac{3}{4}\ln (4x^2+4x+13)+c$

$⟹\lambda =\frac{-3}{4}$

$⟹-4\lambda =3$