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integrate using u substitution
Integrate via substitution $\int \frac{3x+6}{x^2+4x-3}$
Step 1: $u= x^{2}+4x-3$
Step 2: next, find du/dx. If $u=x^2+4x-3$ then $du/dx=2x+4$
Step 3: looks like $\frac{3}{2}\frac{du}{dx}=3x+6$
Step 4: so, $dx=\frac{3 du}{2(3x+6)}$
Step 5: now, put the u and the dx into the original integral to get $\int\frac{3(3x+6)du}{2u(3x+6)}$
Step 6: this simplifies to $\frac{3}{2}\int\frac{du}{u}$
Next step? Make a sketch