Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am having trouble with a question really need help please.

$$ \int {\sqrt{x^2 + 81} \over 2} \,dx $$

I thought about taking the square root off and turning the question into $\frac 12 \int (x^2 +81)^{1/2}\, dx$ but then wondered if I could use the quotient rule.

share|improve this question
Unfortunately, there is no such thing as a "quotient rule" for integration. What other techniques do you know? If this is HW for a course, what section are you currently studying? There is a particular technical method that works, based on the fact that "$x^2 + a^2$" shows up inside a radical. –  Shaun Ault Mar 7 '12 at 13:48
add comment

5 Answers

Hint: Put $x = 9 \tan{t}$, then $x^{2}+81 = 81(\tan^{2}(t)+1) = 81 \cdot \sec^{2}(t)$.

share|improve this answer
+1, very good substitution. –  Américo Tavares Mar 7 '12 at 22:33
add comment

$$ \int \sqrt{x^2+81} dx $$

Integration by parts formula

$$ \int u(x)v^{'}(x) dx = u(x)v(x) - \int v(x)u^{'}(x) dx$$

Therefore assume that $v^{'}(x) = 1$ in this case

Denote the integral

$$ I = \int \sqrt{x^2+81} dx$$

$$ \begin{align*} I &= x \sqrt{x^2+81} - \int \frac{x^2}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - I + 81 \int \frac{1}{\sqrt{x^2+81}} dx \end{align*} $$


$$ 2I = x \sqrt{x^2+81} +81 \int \frac{1}{\sqrt{x^2+81}} dx$$

the rest you should do it yourself.

I was told that we are not supposed to give complete solution for homework questions.


$u(x) = \sqrt{x^2+81}$ and

$v(x) = x$, therefore $v^{'}(x) = 1$

$$ u^{'}(x) = \frac{1}{2}\left(x^2+81\right)^{-\frac{1}{2}} \times 2x = \frac{x}{\sqrt{x^2+81}}$$

$$ \begin{align*} \int \sqrt{x^2+81} dx &= \int u(x) v^{'}(x) \\ &= u(x)v(x) - \int u^{'}(x) v(x) dx\\ &= x \sqrt{x^2+81} - \int \frac{x^2+81-81}{\sqrt{x^2+81}} dx\\ &= x \sqrt{x^2+81} - \int \sqrt{x^2+81} \hspace{3pt} dx + \int \frac{81}{\sqrt{x^2+81}} dx\\ \end{align*} $$

share|improve this answer
In your final expression, the variable $x$ appears outside the integral, without any proper definition there. You should change this notation. –  Martin Mar 7 '12 at 15:12
add comment

The simplest method I see here is to replace x by something else. We know that $a^2+x^2 = x^2+81 = 9^2+x^2$ therefore we can put $x = 9\tan(\theta)$ and get

$dx = 9\sec^2\theta d\theta$

Therefore $$1/2 \int \sqrt{9^2+9^2\tan^2(\theta)} 9\sec^2 \theta d\theta =$$

$$27/2 \int \sqrt{1+\tan^2(\theta)} \sec^2 \theta d\theta =$$

$$27/2 \int \sec^3 \theta d\theta $$

share|improve this answer
you don't need to delete your own answer if someone answers at the same time as you. –  Harry Stern Mar 7 '12 at 18:56
@HarryStern: It is usually a good idea to try to reduce noise, though. –  Aryabhata Mar 7 '12 at 21:14
okay! thanks. PS sec^3 is a lot of fun to integrate –  Sam Creamer Mar 9 '12 at 16:58
I am just new here, but I wonder how people vote here because this elaborates first answer (by Chandrasekhar), but that got 11 votes and this has 0 (I am voting this one)- Weird voting system –  Jeremy Carlos Mar 17 '12 at 15:02
add comment

Added. Euler substitution

  • The Euler substitution $$\begin{equation*} \sqrt{x^{2}+81}=t-x\Leftrightarrow x=\frac{1}{2}\frac{t^{2}-81}{t}, dx=\frac{1}{2}\frac{t^{2}+81}{t^{2}}\;\mathrm{d}t \end{equation*}$$ reduces the given integral to an integral of a rational function in $t$ $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx &=&\frac{1}{2}\int \left( t-\frac{1}{2}\frac{t^{2}-81}{t}\right) \cdot \frac{1}{2}\frac{t^{2}+81}{t^{2}}dt \\ &=&\frac{1}{2}\int \frac{6561+162t^{2}+t^{4}}{4t^{3}}dt=\frac{1}{16}t^{2}-\frac{6561}{ 16t^{2}}+\frac{81}{4}\ln t. \end{eqnarray*}$$

Added 2. Hyperbolic substitution

  • The hyperbolic substitution $$\begin{equation*} x=9\sinh t\Leftrightarrow t=\operatorname{arcsinh}\frac{x}{9},dx=9\cosh tdt,\end{equation*}$$ gives $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx &=&\frac{1}{2}\int 9\sqrt{81\sinh ^{2}t+81} \cosh t\,dt \\ &=&\frac{81}{2}\int \cosh ^{2}t\,dt \\ &=&\frac{81}{2}\int \left( \frac{e^{2t}}{4}+\frac{1}{2}+\frac{e^{-2t}}{4} \right) \,dt \\ &=&\frac{81}{2}\left( \frac{1}{8}e^{2t}+\frac{1}{2}t-\frac{1}{8} e^{-2t}\right). \end{eqnarray*}$$ Since $t=\operatorname{arcsinh}\frac{x}{9}=\ln \left( \frac{x}{9}+\frac{1}{9}\sqrt{x^{2}+81}\right) $, we obtain $$\begin{eqnarray*} \frac{1}{2}\int \sqrt{x^{2}+81}dx&=&\frac{81}{2}\left( \frac{1}{8}e^{2\operatorname{arcsinh}\frac{x}{9}}+\frac{1}{2}\operatorname{arcsinh}\frac{x}{9}-\frac{1}{8}e^{-2\operatorname{arcsinh}\frac{x}{9}}\right) \\ &=&\frac{81}{16}\left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) ^{2}+ \frac{81}{4}\ln \left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) \\ &&-\frac{81}{16}\left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) ^{-2} \\ &=&\frac{81}{4}\ln \left( \frac{1}{9}x+\frac{1}{9}\sqrt{x^{2}+81}\right) +\frac{1}{4}x\sqrt{x^{2}+81}+\text{Constant}. \end{eqnarray*}$$

There is a general method for integrating a function of the form $P(x)/\sqrt{ax^{2}+bx+c}$ I have already posted here.

If $P(x)$ is a polynomial of degree $n\geq 2$, we can find a polynomial $Q(x) $ of degree $n-1$ and a constant $C$ such that$^1$

$$\int \frac{P(x)}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x=Q(x)\sqrt{ax^{2}+bx+c}+\int \frac{C}{\sqrt{ax^{2}+bx+c}}\;\mathrm{d}x.$$

The given integral is $1/2$ of $$ \begin{equation*} I(x):=\int \sqrt{x^{2}+81}dx=\int \frac{x^{2}+81}{\sqrt{x^{2}+81}}\mathrm{d}x=Q(x) \sqrt{x^{2}+81}+\int \frac{C}{\sqrt{x^{2}+81}}\mathrm{d}x, \end{equation*}$$ with $Q(x)=Ax+B$. To find the constants $A$ and $B$ differentiate both sides $$\begin{eqnarray*} \frac{x^{2}+81}{\sqrt{x^{2}+81}} &=&A\sqrt{x^{2}+81}+\frac{2x\cdot \left( Ax+B\right) }{2\sqrt{x^{2}+81}}+\frac{C}{\sqrt{x^{2}+81}} \\ &=&\frac{2Ax^{2}+Bx+81A+C}{\sqrt{x^{2}+81}} \end{eqnarray*}$$ and equate the coefficients in the numerators $$\begin{eqnarray*} 2A &=&1\Leftrightarrow A=\frac{1}{2} \\ B &=&0 \\ 81A+C &=&81\Leftrightarrow C=\frac{81}{2}. \end{eqnarray*}$$ Consequently $$ \begin{equation*} I(x)=\frac{1}{2}x\sqrt{x^{2}+81}+\frac{81}{2}\int \frac{1}{\sqrt{x^{2}+81}}\mathrm{d}x.\end{equation*}$$ Note: This result is essentially the same as Kirthi Raman's. The integral on the right can be easily evaluated using the substitution $x=9u$, because then becomes a direct integral $$\begin{eqnarray*} \int \frac{1}{\sqrt{x^{2}+81}}dx &=&\int \frac{1}{\sqrt{u^{2}+1}}\mathrm{d}u \\ &=&\operatorname{arcsinh} u=\operatorname{arcsinh}\frac{x}{9}+\text{Constant}. \end{eqnarray*}$$

To write it in terms of the natural logarithmic use the identity $$\begin{equation*} \operatorname{arcsinh}u=\ln \left( u+\sqrt{u^{2}+1}\right) \end{equation*}.$$

$^1$ Described in Cálculo Integral em $\mathbb{R}$ by M. Olga Baptista.

share|improve this answer
add comment

There exists a simple formula for the integrals of the type $\int{\sqrt{x^2 + a^2}} dx$ i.e. $$\int{\sqrt{x^2 + a^2}} dx=\space\frac{x}{2}\sqrt{x^2 + a^2}+\frac{a^2}{2}\arctan{\frac{x}{a}}+ C$$ You can prove this by putting $x= a\tan\theta$, differantiating both sides to get $dx=a \ \sec^2\theta \space d\theta$, and evaluating the simple integral that follows. Using this formula, your problem is given by $$\int {\sqrt{x^2 + 81} \over 2} \,dx= \int \sqrt{x^2 + 9^2} \,dx=\frac{x}{2}\sqrt{x^2 + 9^2}+\frac{81}{2} \arctan{\frac{x}{9}+C}$$ Hope this helps.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.