Use integration by parts to prove an integral equation

Let
$I_n (x) = \int_0^x t^n (t^2 + a^2)^{-\frac{1}{2}} \, dt.$

Use integration by parts to prove

$nI_n (x) = x^{n-1} \sqrt{x^2+a^2} - (n-1)a^2 I_{n-2} (x) \qquad \text{if } n \geq 2.$
Using part (a), prove that
$\int_0^2 x^5 (x^2 + 5)^{-\frac{1}{2}} \, dx = \frac{168}{5} - \frac{40 \sqrt{5}}{3}.$

Proof. First, to apply integration by parts define
$\begin{align*} u &= t^{n-1} & \implies && du &= (n-1)^{n-2} \, dt \\ dv &= t(t^2+a^2)^{-\frac{1}{2}} \, dt & \implies && v &= (t^2+a^2)^{\frac{1}{2}}. \end{align*}$

Then, we apply integration by parts to compute,

$\begin{align*} I_n (x) &= \int_0^x t^n (t^2+a^2)^{-\frac{1}{2}} \, dt \\ &= \int_0^x u \, dv \\ &= uv \Bigr \rvert_0^x - \int_0^x v \, du \\ &= t^{n-1} (t^2+a^2)^{\frac{1}{2}} \Bigr \rvert_0^x - \int_0^x (n-1)t^{n-2} (t^2 + a^2)^{\frac{1}{2}} \, dt \\ &= x^{n-1}(x^2+a^2)^{\frac{1}{2}} - \int_0^x (n-1)t^{n-2} (t^2+a^2)(t^2+a^2)^{-\frac{1}{2}} \, dt \\ &= x^{n-1} (x^2+a^2)^{\frac{1}{2}} - \int_0^x (n-1) t^n (t^2+a^2)^{-\frac{1}{2}} \, dt - \int _0^x (n-1) t^{n-2} a^2 (t^2+a^2)^{-\frac{1}{2}} \, dt \\ &= x^{n-1} (x^2+a^2)^{\frac{1}{2}} - (n-1)I_n (x) - a^2(n-1)I_{n-2} (x). \end{align*}$

Thus, adding $(n-1)I_n(x)$ to both sides we have

$nI_n (x) = x^{n-1} (x^2+a^2)^{\frac{1}{2}} - a^2(n-1)I_{n-2} (x). \qquad \blacksquare$
Proof. First from the definition of $I_n (x)$ , letting $a^2 = 5$ we have
$\begin{align*} I_5 (2) &= \int_0^2 t^5 (t^2 + 5)^{-\frac{1}{2}} \, dt \\ I_3 (2) &= \int_0^2 t^3 (t^2 + 5)^{-\frac{1}{2}} \, dt. \end{align*}$

Then, using the formula in the solution of part (a) we have

$\begin{align*} &&5I_5 (2) &= 2^4 \sqrt{4 + 5} - (4)(5)\cdot I_3 (2) \\ \implies && I_5 (2) &= \frac{48}{5} - 4I_3 (2). \end{align*}$

Then, again using the formula in the solution of part (a) we have

$\begin{align*} &&3I_3 (2) &= 2^2 \sqrt{4+5} - (5)(2) \cdot I_1(2) \\ \implies &&I_3 (2) &= 4 - \frac{10}{3} \cdot I_1(2) \\ &&&= 4 - \frac{10}{3} \int_0^2 t(t^2+5)^{-\frac{1}{2}} \, dt. \end{align*}$

Now, to evaluate this integral we use the method of substitution, letting $u = t^2+5$ implies $du = 2t \, dt$ . So,

$\begin{align*} \int_0^2 t(t^2+5)^{-\frac{1}{2}} \, dt &= \frac{1}{2} \int_{u(0)}^{u(2)} u^{-\frac{1}{2}} \, du \\ &= \left. \frac{1}{2} 2u^{\frac{1}{2}} \right|_5^9 \\ &= 3 - \sqrt{5}. \end{align*}$

So, plugging this solution back into our formula for $I_3(2)$ we have

$I_3 (2) = 4 - \frac{10}{3} (3 - \sqrt{5}) = \frac{10}{3} \sqrt{5} - 6.$

Then, using this solution in the formula for $I_5(2)$ we have

$I_5 (2) = \frac{48}{5} - 4 \left( \frac{10}{3} \sqrt{5} - 6 \right) = \frac{168}{5} - \frac{40\sqrt{5}}{3}. \qquad \blacksquare$