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[tex]\int\frac{x}{x-6}dx[/tex]

[tex]u=x-6[/tex]

[tex]\int \frac{u+6}{u}du[/tex]

[tex]\int 1+\frac{6}{u} du[/tex]

[tex]u+6ln|u|+C[/tex]

[tex]x-6+6ln|x-6|+C[/tex]

this appears to be incorrect although it seems logical, also if somone could please tell me the syntax for the definte integral

also similar case here

[tex]\int \frac{x^2}{x+4}dx[/tex]

[tex]u=x+4[/tex]

[tex]\int \frac{(u-4)^2}{u}du[/tex]

[tex]\int \frac{u^2-8u+16}{u}du[/tex]

[tex]\int u-8+\frac{16}{u} du[/tex]

[tex]\frac{u^2}{2}-8u+16ln|u|+C[/tex]

[tex]\frac{(x+4)^2}{2}-8(x+4)+16ln|x+4|+C[/tex]

[tex]u=x-6[/tex]

[tex]\int \frac{u+6}{u}du[/tex]

[tex]\int 1+\frac{6}{u} du[/tex]

[tex]u+6ln|u|+C[/tex]

[tex]x-6+6ln|x-6|+C[/tex]

this appears to be incorrect although it seems logical, also if somone could please tell me the syntax for the definte integral

also similar case here

[tex]\int \frac{x^2}{x+4}dx[/tex]

[tex]u=x+4[/tex]

[tex]\int \frac{(u-4)^2}{u}du[/tex]

[tex]\int \frac{u^2-8u+16}{u}du[/tex]

[tex]\int u-8+\frac{16}{u} du[/tex]

[tex]\frac{u^2}{2}-8u+16ln|u|+C[/tex]

[tex]\frac{(x+4)^2}{2}-8(x+4)+16ln|x+4|+C[/tex]

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