Evaluate $\int (x^{4}+81)/\sqrt{x^{2}+9}\,\mathrm dx$

Question

I need small help in evaluating $$\int\left(\frac{x^{4}+81}{\sqrt{x^{2}+9}}\right)dx$$

My Approach

Let’s assume

$$I=\int\left(\frac{x^{4}+81}{\sqrt{x^{2}+9}}\right) dx$$

$$\implies I=\int\left(\frac{(x^{2}+9)^{2}-18x^{2}}{\sqrt{x^{2}+9}}\right)dx$$

$$\implies I=\int (x^{2}+9)^{\frac{3}{2}}dx-18\int\left(\frac{x^{2}}{\sqrt{x^{2}+9}}\right)dx$$

$$\implies I=\int (x^{2}+9)^{\frac{3}{2}}dx-18\int\left(\frac{(x^{2}+9)-9}{\sqrt{x^{2}+9}}\right)dx$$

$$\implies I=\int({x^{2}+9})^{\frac{3}{2}}dx-18\int\left(\sqrt{x^{2}+9}\right)dx+162\int\left(\frac{1}{\sqrt{x^{2}+9}}\right)dx$$

Let's Assume $$I_{1}=\int({x^{2}+9})^{\frac{3}{2}}dx$$ $$I_{2}=\int\left(\sqrt{x^{2}+9}\right)dx$$ $$I_{3}=\int\left(\frac{1}{\sqrt{x^{2}+9}}\right)dx$$

Case-$1$:

For Evaluating $$I_{3}=\int\left(\frac{1}{\sqrt{x^{2}+9}}\right)dx$$ I just have to substitute $x=3\tan\theta$ and finally the integral will become $$I_{3}=\int\left(\frac{3\sec^{2}\theta}{3\sec\theta}\right)d\theta$$ $$\implies I_{3}=\int(\sec\theta)d\theta$$ $$\implies I_{3}=\ln|\sec\theta+\tan\theta|+C_{1}$$

My Doubt

I really don't know how to evaluate the integrals $$I_{1}=\int(x^{2}+9)^{\frac{3}{2}}dx$$ and $$I_{2}=\int\left(\sqrt{x^{2}+9}\right)dx$$

Please Help me out with Evaluating $I_{1}$ and $I_{2}$

Answer 1

Adding to the comment, If you don't want to procced using Trigonometric substitutions Then, Taken $$ I := \int_{0}^{1} \frac{x^{4}+7}{\sqrt{x^{2}+9}} \,\mathrm dx$$

Let $\displaystyle I_1 := \int_{0}^{1} \frac{\mathrm dx}{\sqrt{x^{2}+9}}$ and $\displaystyle I_2 := \int_{0}^{1} \frac{x^{4}}{\sqrt{x^{2}+9}} \,\mathrm dx$, then $I = 7I_1 + I_2$.

I will leave the evaluation of $I_1$ to you.

Let’s evaluate $I_2$:

Rationalising the integrand we have $$\left(\frac{x^{4}}{{x^{2}+9}}\sqrt{x^{2}+9}\right) =\left( x^2 - 9 + 81\left(\frac{1}{{x^{2}+9}}\right)\right)\sqrt{x^{2}+9}.$$ Now we have

$$I_2 = \int_{0}^{1}x^2\sqrt{x^{2}+9}\,\mathrm dx - 9\int_{0}^{1} \sqrt{x^{2}+9}\,\mathrm dx + 81I_1$$

For the integral $\int_{0}^{1}x^2\sqrt{x^{2}+9}\,\mathrm dx$ you can use the substitution $u = x^{2}+9$ for simplification.

I will leave the evaluation of $\int_{0}^{1} \sqrt{x^{2}+9}\,\mathrm dx$ and rest of the things to you.

Answer 2

Substitute $t=\sec(u)$; $\mathrm dt= \sec(u) \tan(u) \,\mathrm du$.

The integral becomes $\displaystyle \int {\sec(u)(\tan(u))^4} \,\mathrm du$.

Write $\tan^2(u)$ = $\sec^2(u)-1$.

$\displaystyle \int {\sec(u) \cdot (\sec^2(u)-1)^2} \,\mathrm du$.

Expand the binomial and apply reduction formula to get the integral. (I’ll leave it to you; in case you get stuck, let me know and I’ll edit my answer with the detailed steps.)

In case you don’t know the reduction formula for $\sec(x)$ here it is:

$\displaystyle \int {\sec^n(x)} \,\mathrm dx = \frac{\sec^{n-2}(x) \tan(x)}{n-1} +\frac{n-2}{n-1} \int \sec^{n-2}(x)\,\mathrm dx$.

Another way is “guessing”/“assuming” the integral which is applicable when there is a square root of a quadratic in the denominator. Let me know if you want me to explain it briefly or in detail.