Integral equation with a square root seems to go on $$ \int_0^1 \sqrt{1+y^2} (a_0 + a_1y) dy 
$$
I have tried to solve this equation by using integration by parts. I am using the square root as the part which I am differentiating. By using this, the method seems like it is ongoing and does not have a final answer. Any guidance will be much appreciated
I am still trying to figure out the formatting so please forgive me 
 A: $$I=\int_0^1(a_0+a_1x)\sqrt{1+x^2}\ \mathrm dx$$
$$I=a_0\int_0^1\sqrt{1+x^2}\ \mathrm dx+a_1\int_0^1x\sqrt{1+x^2}\ \mathrm dx$$
$$I=a_0I_0+a_1I_1$$

$$I_0=\int_0^1\sqrt{1+x^2}\ \mathrm dx$$
For this we will use a trigonometric substitution:
$$x=\tan u\Rightarrow \mathrm dx=\sec^2u\ \mathrm du$$
The change of bounds:
$$\int_0^1\mapsto \int_0^{\pi/4}$$
Hence
$$I_0=\int_0^{\pi/4}\sqrt{1+\tan^2u}\sec^2u\ \mathrm du$$
$$I_0=\int_0^{\pi/4}\sqrt{\sec^2u}\sec^2u\ \mathrm du$$
$$I_0=\int_0^{\pi/4}\sec u\sec^2u\ \mathrm du=\int_0^{\pi/4}\sec^3u\ \mathrm du$$
We then integrate by parts:
$$\mathrm dV=\sec^2u\ \mathrm du\Rightarrow V=\tan u$$
$$U=\sec u\Rightarrow \mathrm dU=\sec u\tan u\ \mathrm du$$
Which gives
$$I_0=\sec u\tan u\big|_0^{\pi/4}-\int_0^{\pi/4}\tan u\sec u\tan u\ \mathrm du$$
$$I_0=\sqrt{2}-\int_0^{\pi/4}\sec u\tan^2u\ \mathrm du$$
$$I_0=\sqrt{2}-\int_0^{\pi/4}\sec u(\sec^2u-1)\ \mathrm du$$
$$I_0=\sqrt{2}-\int_0^{\pi/4}\sec^3u\ \mathrm du+\int_0^{\pi/4}\sec u\ \mathrm du$$
$$I_0=\sqrt{2}-I_0+\int_0^{\pi/4}\sec u\ \mathrm du$$
$$2I_0=\sqrt{2}+\int_0^{\pi/4}\sec u\ \mathrm du$$
$$2I_0=\sqrt{2}+\log(\sec u+\tan u)\big|_0^{\pi/4}$$
$$2I_0=\sqrt{2}+\log(1+\sqrt2)$$
$$I_0=\frac{\sqrt{2}+\log(1+\sqrt2)}2$$

$$I_1=\int_0^1x\sqrt{1+x^2}\ \mathrm dx$$
Substitution:
$$w^2=x^2+1\Rightarrow w\mathrm dw=x\mathrm dx$$
$$\int_0^1\mapsto \int_1^{\sqrt2}$$
This gives
$$I_1=\int_1^{\sqrt2}\sqrt{w^2}w\ \mathrm dw$$
$$I_1=\int_1^{\sqrt2}w^2\ \mathrm dw$$
$$I_1=\frac{w^3}3\bigg|_1^{\sqrt2}$$
$$I_1=\frac{2^{3/2}-1}3$$

Finally:
$$I=\frac{a_0}2\bigg(\sqrt2+\log(1+\sqrt2)\bigg)+\frac{a_1}3\big(2^{3/2}-1\big)$$
