(a):
I = β« x x 2 + 1 d x I=\int \dfrac{x}{\sqrt{x^2+1}}dx I = β« x 2 + 1 β x β d x
I = β« x x 2 + 1 d x I=\int \dfrac{x}{\sqrt{x^2+1}}dx I = β« x 2 + 1 β x β d x
Put x 2 + 1 = t x^2+1=t x 2 + 1 = t
β 2 x d x = d t β x d x = d t 2 \Rightarrow 2xdx=dt
\\\Rightarrow xdx=\dfrac{dt}{2} β 2 x d x = d t β x d x = 2 d t β
So, I = 1 2 β« 1 t d t I=\dfrac{1}{2}\int \dfrac{1}{\sqrt{t}}dt I = 2 1 β β« t β 1 β d t
I = 1 2 β« 1 t d t I=\dfrac{1}{2}\int \dfrac{1}{\sqrt{t}}dt I = 2 1 β β« t β 1 β d t
I = 1 2 β« t β 1 / 2 d t = 1 2 . t 1 / 2 1 / 2 + c = t + c = x 2 + 1 + c I=\dfrac{1}{2}\int t^{-1/2}dt
\\=\dfrac{1}{2}.\dfrac{t^{1/2}}{1/2}+c
\\=\sqrt t+c
\\=\sqrt{x^2+1}+c I = 2 1 β β« t β 1/2 d t = 2 1 β . 1/2 t 1/2 β + c = t β + c = x 2 + 1 β + c
I = 1 2 β« t β 1 / 2 d t = 1 2 . t 1 / 2 1 / 2 + c = t + c = x 2 + 1 + c I=\dfrac{1}{2}\int t^{-1/2}dt
\\=\dfrac{1}{2}.\dfrac{t^{1/2}}{1/2}+c
\\=\sqrt t+c
\\=\sqrt{x^2+1}+c I = 2 1 β β« t β 1/2 d t = 2 1 β . 1/2 t 1/2 β + c = t β + c = x 2 + 1 β + c I = 1 2 β« t β 1 / 2 d t = 1 2 . t 1 / 2 1 / 2 + c = t + c = x 2 + 1 + c I=\dfrac{1}{2}\int t^{-1/2}dt
\\=\dfrac{1}{2}.\dfrac{t^{1/2}}{1/2}+c
\\=\sqrt t+c
\\=\sqrt{x^2+1}+c I = 2 1 β β« t β 1/2 d t = 2 1 β . 1/2 t 1/2 β + c = t β + c = x 2 + 1 β + c
(b):
The graph using matlab from (-5,5) is as follows:
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