Consider the following example:

$\mathrm{eq1}≔p\left(x\right)-\frac{1}{2}\left(\mathrm{Int}\left(xyp\left(y\right)\,y=0..1\right)\right)=\frac{5}{6}x$

$\mathrm{eq1}≔p\left(x\right)-\frac{\left({\int }_0^{1}xyp\left(y\right)\ⅆy\right)}{2}=\frac{5x}{6}$

$\mathrm{intsolve}\left(\mathrm{eq1}\,p\left(x\right)\right)$

$p\left(x\right)=x$

If we instead use the 'Neumann' method, an approximation to this result is returned:

$\mathrm{intsolve}\left(\mathrm{eq1}\,p\left(x\right)\,\mathrm{method}=\mathrm{Neumann}\right)$

$p\left(x\right)=\frac{279935x}{279936}$

$\mathrm{intsolve}\left(\mathrm{eq1}\,p\left(x\right)\,\mathrm{method}=\mathrm{Neumann}\,\mathrm{order}=20\right)$

$p\left(x\right)=\frac{21936950640377855x}{21936950640377856}$

Computing the solution and verifying the result:

$\mathrm{eq2}≔f\left(x\right)=x+1+\mathrm{Int}\left(\left(1+2\left(x-y\right)\right)f\left(y\right)\,y=0..x\right)$

$\mathrm{eq2}≔f\left(x\right)=x+1+{\int }_0^x\left(1+2x-2y\right)f\left(y\right)\ⅆy$

$\mathrm{intsolve}\left(\mathrm{eq2}\,f\left(x\right)\right)$

$f\left(x\right)={\ⅇ}^{2x}$

To verify this solution, transform it first into a procedure:

$f=\mathrm{unapply}\left(\mathrm{rhs}\left(\right)\,x\right)$

$f=\left(x\mapsto {\ⅇ}^{2\cdot x}\right)$

$\mathrm{eval}\left(\mathrm{eq2}\,\right)$

${\ⅇ}^{2x}=x+1+{\int }_0^x\left(1+2x-2y\right){\ⅇ}^{2y}\ⅆy$

$\mathrm{value}\left(\right)$

${\ⅇ}^{2x}={\ⅇ}^{2x}$

For some type of problems the solution obtained using the Laplace method contains Dirac functions:

$\mathrm{eq3}≔\mathrm{Int}\left(\exp \left(a\left(x-y\right)\right)f\left(y\right)\,y=0..x\right)=1$

$\mathrm{eq3}≔{\int }_0^x{\ⅇ}^{a\left(x-y\right)}f\left(y\right)\ⅆy=1$

$\mathrm{intsolve}\left(\mathrm{eq3}\,f\left(x\right)\,\mathrm{method}=\mathrm{Laplace}\right)$

$f\left(x\right)=\mathrm{Dirac}\left(x\right)-a$

Another type of integral equation automatically solved using the eigenfunction approach:

$\mathrm{eq4}≔f\left(x\right)+b\left(\mathrm{Int}\left(\left(xy+x^2{y}^2\right)f\left(y\right)\,y=-1..1\right)\right)=d\left(x\right)$

$\mathrm{eq4}≔f\left(x\right)+b\left({\int }_{\mathrm{-1}}^1\left(x^2{y}^2+xy\right)f\left(y\right)\ⅆy\right)=d\left(x\right)$

$\mathrm{intsolve}\left(\mathrm{eq4}\,f\left(x\right)\right)$

$f\left(x\right)={\int }_{\mathrm{-1}}^1\left(-\frac{5x^{2}b{y}^2}{2b+5}-\frac{3xby}{2b+3}\right)d\left(y\right)\ⅆy+d\left(x\right)$

An example where an arbitrary constant $\mathrm{\_C1}$ is introduced with the solution:

$\mathrm{eq5}≔3x^2+4x=\mathrm{Int}\left(\left(6x^{2}y+4x{y}^2\right)p\left(y\right)\,y=-1..1\right)$

$\mathrm{eq5}≔3x^2+4x={\int }_{\mathrm{-1}}^1\left(6x^{2}y+4x{y}^2\right)p\left(y\right)\ⅆy$

$\mathrm{intsolve}\left(\mathrm{eq5}\,p\left(x\right)\right)$

$p\left(x\right)=x^2\mathrm{\_C1}+\frac{3}{4}x+\frac{3}{2}-\frac{3}{5}\mathrm{\_C1}$

An example which is in the form of an integral transform:

$\mathrm{eq6}≔\mathrm{Int}\left(\frac{f\left(y\right)}{1+y^2}\sin \left(xy\right)\,y=0..\mathrm{\infty }\right)=\exp \left(-3x\right)$

$\mathrm{eq6}≔{\int }_0^{\mathrm{\infty }}\frac{f\left(y\right)\sin \left(xy\right)}{y^2+1}\ⅆy={\ⅇ}^{-3x}$

$\mathrm{intsolve}\left(\mathrm{eq6}\,f\left(x\right)\right)$

$f\left(x\right)=\frac{2x\left(x^2+1\right)}{\mathrm{\pi }\left(x^2+9\right)}$

The solution of a Fredholm integral equation can be approximated using the collocation method:

$a≔0\:$

$b≔5\:$

$g≔x\mapsto -x\:$

$k≔\left(x\&comma;y\right)\mapsto \mathrm{piecewise}\left(x<y\&comma;x\cdot \left(b-y\right)\&comma;y\cdot \left(b-x\right)\right)\&colon;$

$\mathrm{eq7}≔f\left(x\right)+\mathrm{int}\left(f\left(y\right)k\left(x\&comma;y\right)\&comma;y=a..b\right)+g\left(x\right)$

$\mathrm{eq7}≔f\left(x\right)+{\int }_0^5\left(\left\{\begin{array}{cc}x\left(5-y\right)& x<y\\ y\left(5-x\right)& \mathrm{otherwise}\end{array}\right.\right)f\left(y\right)\&DifferentialD;y-x$

$p≔\mathrm{unapply}\left(\mathrm{rhs}\left(\mathrm{intsolve}\left(\mathrm{eq7}\&comma;f\left(x\right)\&comma;\mathrm{method}=\mathrm{collocation}\&comma;\mathrm{order}=10\right)\right)\&comma;x\right)$

$p≔x\mapsto \frac{220000}{8122882553}\cdot x^{10}-\frac{33081098947000}{60853654989179049}\cdot x^9+\frac{98331769350000}{20284551663059683}\cdot x^8-\frac{501031314879600}{20284551663059683}\cdot x^7+\frac{1594256521707500}{20284551663059683}\cdot x^6-\frac{6502492915408455}{40569103326119366}\cdot x^5+\frac{8429866033778125}{40569103326119366}\cdot x^4-\frac{9902325865977605}{60853654989179049}\cdot x^3+\frac{1401159656806875}{20284551663059683}\cdot x^2-\frac{231864099964442}{20284551663059683}\cdot x$

Plot of the approximate solution:

$\mathrm{plot}\left(p\&comma;a..b\&comma;\mathrm{title}=''Approximate\; Solution\; of\; Integral\; Equation''\&comma;\mathrm{color}=\mathrm{blue}\right)$

To test how close the approximation matches the actual solution, we can sample the absolute value of the expression for the integral equation over the interval:

$n≔100\&colon;$

$q≔\mathrm{unapply}\left(\mathrm{abs}\left(\mathrm{eval}\left(\mathrm{eq7}\&comma;f=p\right)\right)\&comma;x\right)\&colon;$

$X≔\mathrm{Vector}\left(n\&comma;i\mapsto \mathrm{evalf}\left(a+\frac{\left(i-1\right)\cdot \left(b-a\right)}{n-1}\right)\right)\&colon;$

$E≔\mathrm{`~`}\left[q\right]\left(X\right)\&colon;$

$\mathrm{dataplot}\left(X\&comma;E\&comma;\mathrm{style}=\mathrm{line}\&comma;\mathrm{title}=''Error\; of\; Approximate\; Solution\; of\; Integral\; Equation''\right)$

$\mathrm{avg\_error}≔\frac{\mathrm{add}\left(E\right)}{n}$

$\mathrm{avg\_error}≔0.0001670862960$

$\mathrm{max\_error}≔\max \left(E\right)$

$\mathrm{max\_error}≔0.001720013949$

The Laplace method can be used to solve singular integral equations:

$\mathrm{eq8}≔\mathrm{Int}\left(\frac{f\left(y\right)}{\mathrm{sqrt}\left(x-y\right)}\&comma;y=0..x\right)-x^2$

$\mathrm{eq8}≔{\int }_0^x\frac{f\left(y\right)}{\sqrt{x-y}}\&DifferentialD;y-x^2$

$\mathrm{intsolve}\left(\mathrm{eq8}\&comma;f\left(x\right)\&comma;\mathrm{method}=\mathrm{Laplace}\right)$

$f\left(x\right)=\frac{8x^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{3\mathrm{\pi }}$