Antiderivative of $x\sqrt{1+x^2}$

Question

I am attempting this problem given to me, but the answer key does not explain the answer. The question asks me to find the antiderivative of $x\sqrt{1+x^2}$.

My attempt:

$\frac{d}{dx}f(g(x)) = g'(x)f'(g(x))$

$g'(x)$ must be $x$, therefore $g(x)=\frac{1}{2}x^2$

$f'(x)$ must equal $\sqrt{1+x^2}$, therefore $f(x)=\frac{2}{3}(1+x^2)^{\frac{3}{2}}$

But then I saw that if $g(x)=\frac{1}{3}x^2$, then $g'(x)=\frac{2}{3}x$, and $f(x)=(1+x^2)^{\frac{3}{2}}$, then $\frac{d}{dx}f(g(x))'=g'(x)f'(g(x))=(2/3)x(3/2)\sqrt{1+x^2}=x\sqrt{1+x^2}$.

But I saw this by chance; what would be a more general and effective way to do this? Should I check $this$, and then write $that$, and put together $those$, etc.? Or is the way I just 'saw' the way the numbers should fall together how one would normally approach this?

Answer 1

You're effectively doing a $u$-substitution to compute $$\int{x\sqrt{1+x^2} \,\mathrm{d}x}$$ with $u=1+x^2$ and hence $du=2xdx$. This turn the earlier integral into $$\int{\frac{1}{2}u^{1/2}\,\mathrm{d}u}$$ We can easily integrate this to get $$\frac{2}{3}\frac{1}{2}u^{3/2}+C = \frac{1}{3}(1+x^2)^{3/2}+C$$ $u$-substitution can be thought of as being the integral-version of chain rule, much in the same way that integration by-parts is the integral-version of product rule.

Answer 2

$u$-substitution is more standard, but I can follow your method to see how it works out.

$$\begin{align*} \frac d{dx}f(g(x)) &= g'(x) \cdot f'(g(x))\\ &= x\sqrt{1+x^2} \end{align*}$$

Try $g'(x) = x$, then

$$g(x) = \frac12x^2+C_1$$

Try $C_1 = \frac12$ to get $g(x) = \frac12(1+x^2)$. Then $$\begin{align*} f'(x) &= \sqrt{2x}\\ &= \sqrt2 \cdot \sqrt x\\ f(x) &= \sqrt2\cdot\frac{2}3 x^{3/2}+C\\ f(g(x)) &= \sqrt2\cdot\frac23\left[\frac12(1+x^2)\right]^{\frac32}+C\\ &= \frac13(1+x^2)^{3/2}+C \end{align*}$$

Answer 3

Knowing that a derivative of powers reduces that power by one leads one to consider $$\frac{d}{dx} \, (1 + x^2)^{3/2} = 3 \, x \, \sqrt{1+x^2}.$$ From this it is easily seen that $$\int x \, \sqrt{1 + x^2} \, dx = \frac{1}{3} \, (1 + x^2)^{3/2} + c_{0}.$$

Answer 4

Alternatively: $$\int{x\sqrt{1+x^2} \,\mathrm{d}x} = \frac12\int{\sqrt{1+x^2} \,\mathrm{d}x^2}=\\ \frac12\int \sqrt{1+t} \, \ \mathrm dt=\frac12\cdot \frac23\cdot (1+t)^{3/2}+C=\frac13(1+x^2)^{3/2}+C.$$

Answer 5

Whenever you are finding the integral of a function which consists of f(x) & f'(x) of some function ( Here f(x)=x^2 & f'(x)=2x)

Always go by substituting f(x) to some other variable

Put t^2 = 1+x^2 (I'm taking like this because square root of t^2 will result t which makes my integration easier)

Then 2t dt = 2x dx ==> t dt = x dx

Substitute these values in the integration..

∫x√(1+x^2) dx = ∫√t^2 t dt =∫t×t dt =∫t^2 dt =(t^3)/3

Finally substituting t = √(1+x^2) gives the anti derivative ((1+x^2)^(3/2))/3