Let #160;I= #8747; #160;x+1 #160;2x2+3 #160;dxAlso, #160;x+1= #955;ddx2x2+3+ #956; #8658;x+1= #955;4x+ #956;Equating #160;coefficient ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Special Integrals - 2

∫ x+12 x 2+3 ...

Question

$\int (x + 1) \sqrt{2 x^{2} + 3} d x$

Open in App

Solution

$Let I = \int (x + 1) \sqrt{2 x^{2} + 3} d x Also, x + 1 = λ \frac{d}{d x} (2 x^{2} + 3) + μ \Rightarrow x + 1 = λ (4 x) + μ Equating coefficient of like terms 4 λ = 1 \Rightarrow λ = \frac{1}{4} and μ = 1 ∴ I = \int [\frac{1}{4} (4 x) + 1] \sqrt{2 x^{2} + 3} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 x^{2} + 3} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \int \sqrt{2 (x^{2} + \frac{3}{2})} d x = \frac{1}{4} \int (4 x) \sqrt{2 x^{2} + 3} d x + \sqrt{2} \int \sqrt{x^{2} + {(\frac{\sqrt{3}}{\sqrt{2}})}^{2}} d x Let 2 x^{2} + 3 = t \Rightarrow 4 x d x = d t ∴ I = \frac{1}{4} \int \sqrt{t} d t + \sqrt{2} [\frac{x}{2} \sqrt{x^{2} + \frac{3}{2}} + \frac{3}{4} \log |x + \sqrt{x^{2} + \frac{3}{2}}|] = \frac{1}{4} [\frac{t^{\frac{3}{2}}}{\frac{3}{2}}] + \sqrt{2} \frac{x}{2} \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}} + \frac{3 \sqrt{2}}{4} \log |x + \frac{\sqrt{2 x^{2} + 3}}{\sqrt{2}}| + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\frac{\sqrt{2} x + \sqrt{2 x^{2} + 3}}{\sqrt{2}}| + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\sqrt{2} x + \sqrt{2 x^{2} + 3}| - \frac{3 \sqrt{2}}{4} \log \sqrt{2} + C = \frac{1}{6} {(2 x^{2} + 3)}^{\frac{3}{2}} + \frac{x}{2} \sqrt{2 x^{2} + 3} + \frac{3 \sqrt{2}}{4} \log |\sqrt{2} x + \sqrt{2 x^{2} + 3}| + C' Where C' = C - \frac{3 \sqrt{2}}{4} \log \sqrt{2}$

Suggest Corrections