Mathematics

$$\int \sin^{-1} \ (\dfrac{2x +2}{\sqrt{4x^{2} +8x +13}}\ )dx$$ = 


SOLUTION
$$\dfrac{2x+2}{\sqrt{4x^2+8x+3}}=\dfrac{2(x+1)}{\sqrt{4(x^2(4+1)+3^2}}$$
$$=\dfrac{2(x+1)}{\sqrt{4(x+1)^2+3^2}}$$
set $$x+1=\dfrac{3}{2}\tan\theta$$
$$=\dfrac{2\cdot \dfrac{3}{2}\tan \theta}{\sqrt{4\times \dfrac{9}{4}\tan^2\theta +3^2}}=\dfrac{3\tan\theta}{\sqrt{3^2(1+\tan^2\theta)}}$$
$$=\dfrac{3\tan\theta}{3\sin\theta}=\dfrac{\sin\theta}{\cos\theta}\cdot\cos\theta =\sin \theta$$
$$=\sin^{-1}\left(\dfrac{2x+2}{\sqrt{4x^2+\theta x+13}}\right)=\sin^{-1}(\sin \theta)$$
$$=\theta$$
$$x+1=\dfrac{3}{2}\tan\theta$$
$$dx=\dfrac{3}{2}\sin^2\theta d\theta$$ 
$$\tan\theta =\dfrac{2}{3}(x+1)$$
$$\sin\theta =\sqrt{1+\tan^2\theta}$$
$$=\sqrt{1+\dfrac{4}{9}(x+1)^2}$$
$$=\dfrac{1}{3}\sqrt{4x^2+8x+13}$$
$$I=\displaystyle\int\theta \cdot\dfrac{3}{2}\sin^2\theta\cdot d\theta$$
$$=\dfrac{3}{2}\displaystyle\int \theta\cdot \sec^2\theta\cdot d\theta$$
$$=\dfrac{3}{2}[\theta \cdot \tan \theta-\displaystyle\int \tan\theta d\theta]$$
$$=\dfrac{3}{2}[\theta \tan \theta -ln \sec\theta]+c$$
$$=\dfrac{3}{2}[\dfrac{2}{3}(x+1)\tan \{\dfrac{2}{3}(x+1)\}$$ $$-ln\left(\dfrac{\sqrt{4n^2+dn+13}}{3}\right)+c$$
$$=\dfrac{3}{2}\cdot\dfrac{2}{3}(n+1)\tan^{-1}\left\{\dfrac{2}{3}(n+1)\right\}-\dfrac{3}{2}[ln \sqrt{4n^2+dn+13}-ln 3]+c$$
$$=(x+1)\tan^{-1}\left\{\dfrac{2}{3}(n+1)\right\}-\dfrac{3}{2}\cdot \dfrac{1}{2}ln(4n^2+dn+13)+\dfrac{3}{2}ln 3+c$$ New laws
$$=(x+1)\tan^{-1}\left\{\dfrac{2}{3}(x+1)\right\}-\dfrac{3}{4}ln(4n^2+dn+13)+c^2$$.
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