Question

$\int \frac{3x-4}{\sqrt{2x^2+4x+5}}dx=$

• A. $\frac{3}{2}\sqrt{2x^2+4x+5}+\frac{7}{\sqrt{2}}\sin h^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{3}}\right)-c$
• B. $\frac{2}{3}\sqrt{2x^2-4x-5}+\frac{7}{\sqrt{2}}\sin h^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{3}}\right)+c$
• C. $\frac{3}{2}\sqrt{2x^2+4x+5}-\frac{7}{\sqrt{2}}\sin h^{-1}\left(\frac{\sqrt{2}(x-1)}{\sqrt{3}}\right)-c$
• D. $\frac{3}{2}\sqrt{2x^2+4x+5}-\frac{7}{\sqrt{2}}\sin h^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{3}}\right)+c$

Answer 1

Answer: (D)

$\frac{3}{2}\sqrt{2x^2+4x+5}-\frac{7}{\sqrt{2}}\sin h^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{3}}\right)+c$

$\int \frac{3x+4}{\sqrt{2x^2+4x+5}}dx$
lets write
$N\left(x\right)=\mu \left(D^1\right(x\left)\right)+\lambda$
$3x-4=\mu [4x+4]+\lambda$
$4\mu x=3x\Rightarrow \mu ={}^3{/}_4;\lambda =-7$
$\int \frac{N\left(x\right)}{\sqrt{D\left(x\right)}}dx=\int \frac{\mu D^1\left(x\right)}{\sqrt{D\left(x\right)}}+\int \frac{\lambda }{\sqrt{D\left(x\right)}}dx$
$2\mu \sqrt{D}\left(x\right)+\int \frac{\lambda }{\sqrt{D}\left(x\right)}dx$
$\int \frac{1}{\sqrt{2x^2+4x+5}}dx=\int \frac{1}{\sqrt{2}\sqrt{x^2+2x+1{+}^3{/}_2}}dx={}^1{/}_{\sqrt{2}}\int \frac{dx}{\sqrt{(x+1{)}^2+(\sqrt{{}^3{/}_2})}}$
$=\frac{\sqrt{2}}{\sqrt{3}}si{n}^{-1}\left(\frac{(x+1)}{\sqrt{3}}\sqrt{2}\right)+c$
$=2\left(\frac{3}{4}\right)\sqrt{2x^2+4x+5}-\frac{7}{\sqrt{2}}si{n}^{-1}\left(\frac{\sqrt{2}(x+1)}{\sqrt{3}}\right)+c$