Jtdylktuy

Evaluate $$int frac{x^3}{sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=sqrt {u-1}$ and $dx=frac{1}{2sqrt{u-1}}du$.

Therefore, $$int frac{(u-1)sqrt{u-1}}{sqrt u}frac{1}{2sqrt {u-1}}du$$
$$=frac{1}{2}int {sqrt u}-{frac {1}{sqrt u}du}$$
$$={frac{2}{3}}{{(x^2+1})}^{3/2}-{frac{1}{2}}{sqrt {x^2+1}}+C$$for some constant $C$

You made a small mistake on the last line; integrating $frac{1}{2}u^{1/2}-frac{1}{2}u^{-1/2}$ should give $frac{1}{3}u^{3/2}-u^{1/2}+C$, not $frac{2}{3}u^{3/2}-frac{1}{2}u^{1/2}+C$. When you want to double-check an antiderivative, it's worth differentiating it to see if you get the expected integrand.

Here I give you an alternative approach. If we make $x = sinh(u)$, we get

begin{align*}
intfrac{x^{3}}{sqrt{x^{2}+1}}mathrm{d}x & = intsinh^{3}(u)mathrm{d}u = int(cosh^{2}(u)-1)sinh(u)mathrm{d}u\
& = frac{cosh^{3}(u)}{3} - cosh(u) + K
end{align*}

Since $x = sinh(u) = pmsqrt{cosh^{2}(u)-1}$, we conclude that $cosh(u) = sqrt{x^{2} + 1}$. Finally, it results that

begin{align*}
intfrac{x^{3}}{sqrt{x^{2}+1}}mathrm{d}x = frac{(x^{2}+1)^{3/2}}{3} - sqrt{x^{2}+1} + K
end{align*}

Just to give another approach, let $x^2+1=u^2$, in which case $xdx=udu$ and we have

$$int{x^3oversqrt{x^2+1}}dx=int{x^2cdot xdxoversqrt{x^2+1}}=int{(u^2-1)uduover u}=int(u^2-1)du={1over3}u^3-u+C\={1over3}(x^2+1)^{3/2}-(x^2+1)^{1/2}+C$$

The main virtue (if any) of this approach is that it gets rid of the square root symbol for the integration step.

Method 1:

Another approach using trigonometric substitutions

begin{equation}
I =int frac{x^3}{sqrt {x^2+1}}dx
end{equation}

Here let $x = tan(theta)$ we arrive at:

begin{align}
I &=int frac{tan^3(theta)}{sqrt {tan^2(theta)+1}}sec^2(theta):dtheta = int frac{tan^3(theta)}{sec(theta)}sec^2(theta):dtheta\
&= int tan^3(theta)sec(theta):dtheta = int sec(theta)tan(theta) tan^2(theta):dtheta \
&= int sec(theta)tan(theta) left(sec^2(theta) - 1right):dtheta
end{align}

Now let $u = sec(theta)$ to yield:

begin{equation}
int sec(theta)tan(theta)left(u^2 - 1right) frac{:dtheta}{sec(theta)tan(theta)} = int u^2 - 1 :du = frac{u^3}{3} - u + C
end{equation}

Where $C$ is the constant of integration.

Now $u = sec(theta)$ and $x = tan(theta)$ and so $u = sec(arctan(x)) = sqrt{x^2 + 1}$

Thus,

begin{equation}
I =int frac{x^3}{sqrt {x^2+1}}dx = frac{1}{3}left( sqrt {x^2+1}right)^3 - sqrt {x^2+1} + C = frac{1}{3}left(x^2+1right)^{frac{3}{2}} - sqrt {x^2+1} + C
end{equation}

Method 2:

begin{align}
I &=int frac{x^3}{sqrt {x^2+1}}dx = int x cdot frac{x^2}{sqrt {x^2+1}}dx \
&= int x cdot frac{x^2 + 1 - 1}{sqrt {x^2+1}}dx = int x left( sqrt {x^2+1} - frac{1}{sqrt{x^2+1}}right)dx
end{align}

Here let $u = x^2 + 1$:

begin{align}
I &= int x left( sqrt {u} - frac{1}{sqrt{u}}right)frac{du}{2x} = frac{1}{2}left[int left( u^{frac{1}{2}} - u^{-frac{1}{2}}right)duright] = frac{1}{2}left[frac{ u^{frac{3}{2}}}{frac{3}{2}} + frac{ u^{frac{1}{2}}}{frac{1}{2}} right] + C \
&= frac{1}{3}u^{frac{3}{2}} + u^{frac{1}{2}} + C = frac{1}{3}left(x^2 + 1right)^{frac{3}{2}} + left(x^2 + 1right)^{frac{1}{2}} + C
end{align}

Where $C$ is the constant of integration.