evaluating $\int x^2\sqrt{x^2+16}\;\mathrm{d}x$

Question

How do I find $$\int x^2\sqrt{x^2+16} \;\mathrm{d}x$$

I tried a tangent-based substitution, but it didn't seem to work.

Answer 1

Hint: Use the substitution $x=4\sinh(t)$.

Answer 2

Since the integrand is a rational function of $x$ and $\sqrt{x^{2}+a^{2}}$, $a=4$, the two standard substitutions are:

1. the trigonometric substitution $x=a\tan t$

2. the hyperbolic substitution $x=a\sinh t$

For 1. use the trigonometric identity \begin{equation*} 1+\tan ^{2}t=\sec ^{2}t. \end{equation*}

For 2. use the hyperbolic identity \begin{equation*} 1+\sinh ^{2}t=\cosh ^{2}t. \end{equation*}

The 3rd. standard substitution is an algebraic one, the Euler substitution

\begin{equation*} \sqrt{x^{2}+16}=x+t\Leftrightarrow x=\frac{16-t^{2}}{2t},\qquad dx=-\frac{ 16+t^{2}}{2t^{2}}dt, \end{equation*}

which reduces the given integral in $x$ to an integral of a rational function in $t$. We have that \begin{eqnarray*} \int x^{2}\sqrt{x^{2}+16}\,dx &=&-\int \left( \frac{16-t^{2}}{2t}\right) ^{2}\left( \frac{16-t^{2}}{2t}+t\right) \frac{16+t^{2}}{2t^{2}}dt,\qquad t= \sqrt{ x^{2}+16}-x \\ &=&-\int \frac{4096}{t^{5}}-\frac{32}{t}+\frac{1}{16}t^{3}dt \\ &=&\frac{1024}{t^{4}}+32\ln |t|-\frac{t^{4}}{64}+C \\ &=&\frac{1024}{\left( \sqrt{x^{2}+16}-x\right) ^{4}}+32\ln \left| \sqrt{ x^{2}+16}-x\right| -\frac{\left( \sqrt{x^{2}+16}-x\right) ^{4}}{64}+C. \end{eqnarray*}

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ADDED. I assume that the tangent substitution you mention in the question is \begin{equation*} x=4\tan t\Leftrightarrow t=\arctan \frac{x}{4},\qquad dx=(4+4\tan^2 t)dt=4\sec^2 t\,dt, \end{equation*}

which using the trigonometric identity written above gives

\begin{eqnarray*} \int x^{2}\sqrt{x^{2}+16}\,dx &=&\int 16\left( \tan ^{2}t\right) \sqrt{16\tan ^{2}t+16}\left( 4+4\tan ^{2}t\right) \,dt \\ &=&16\times 4\times 4\int \left( \tan ^{2}t\right) \sqrt{\tan ^{2}t+1}\left( 1+\tan ^{2}t\right) \,dt \\ &=&256\int \tan ^{2}t\sec ^{3}t\,dt \\ &=&256\int \frac{\sin ^{2}t}{\cos ^{5}t}\,dt. \end{eqnarray*}

The integrand is a rational function of $\sin t,\cos t$ which is integrable by the Weirstrass substitution \begin{equation*} u=\tan \frac{t}{2},\qquad dt=\frac{2}{\left( 1+u^{2}\right) }\,du. \end{equation*}

Since \begin{equation*} \sin t=\frac{2u}{1+u^{2}},\qquad \cos t=\frac{1-u^{2}}{1+u^{2}}, \end{equation*} we obtain

\begin{eqnarray*} 256\int \frac{\sin ^{2}t}{\cos ^{5}t}\,dt &=&256\int \frac{\left( \frac{2u}{ 1+u^{2}}\right) ^{2}}{\left( \frac{1-u^{2}}{1+u^{2}}\right) ^{5}}\frac{2}{ \left( 1+u^{2}\right) }\,du,\qquad u=\tan \frac{t}{2} \\ &=&512\int \frac{4u^{6}+8u^{4}+4u^{2}}{\left( 1-u\right) ^{5}\left( 1+u\right) ^{5}}du. \end{eqnarray*}

Finally we expand the integrand into partial fractions \begin{eqnarray*} \frac{4u^{6}+8u^{4}+4u^{2}}{\left( 1-u\right) ^{5}\left( 1+u\right) ^{5}} &=&-\frac{1}{2\left( u-1\right) ^{5}}-\frac{3}{4\left( u-1\right) ^{4}}- \frac{3}{8\left( u-1\right) ^{3}}-\frac{1}{16\left( u-1\right) ^{2}}+\frac{1 }{16\left( u-1\right) } \\ &&+\frac{1}{2\left( 1+u\right) ^{5}}-\frac{3}{4\left( 1+u\right) ^{4}}+\frac{ 3}{8\left( 1+u\right) ^{3}}-\frac{1}{16\left( 1+u\right) ^{2}}-\frac{1}{ 16\left( 1+u\right) }. \end{eqnarray*}

Answer 3

As AméricoTavares has found $$\int x^2\sqrt{x^2+16}dx=256\int\tan^2t\sec^3tdt$$

Now, $\displaystyle\tan^2t\sec^3t=(\sec^2t-1)\sec^3t=\sec^5t-\sec^3t$

Now use the reduction formula(proof) of $\displaystyle\int\sec^nt\ dt$ and the terminating case being this

Again as $\displaystyle x=4\tan t\implies t=\arctan\frac x4,$ using this , $\displaystyle-\frac\pi2\le t\le \frac\pi2\implies\sec t>0$ and $\sec t=+\sqrt{1+\left(\frac x4\right)^2}$

Answer 4

$$I=\int x^2\mathrm {\sqrt{x^2+16}}dx$$$$=\int(x^2+16-16)\mathrm {\sqrt{x^2+16}} dx$$$$=\int(x^2+16)\mathrm {\sqrt{x^2+16}}dx-\int16\mathrm {\sqrt{x^2+16}}dx$$$$=\int\mathrm {(x^2+16)}^{\frac{3}{2}}dx-\int 16\mathrm {\sqrt{x^2+16}}.dx$$On solving we get$$I=\frac{1024}{\left( \sqrt{x^{2}+16}-x\right) ^{4}}+32\ln \left| \sqrt{ x^{2}+16}-x\right| -\frac{\left( \sqrt{x^{2}+16}-x\right) ^{4}}{64}+C. $$