kan iemand mij helpen de juiste oplossing zou (3π/2)-3 zijn.
\( \int_{-1}^1 ln(x²+1) \)
\( dv=dx => v=x \)
\( u=ln(x²+1) => du/dx=2x/(x+1) \)
\( <=> du = 2x/(x+1) dx \)
adhv \( \int u dv = uv - \int v du \)
\(=ln(x²+1)x - \int_{-1}^1 3x/(x+1) dx \)
\( =[ln (x²+1)x - 3x/(x²+1)]_{-1}^1 \)
\( = (ln 2 - 3/2) - (-ln 2 - (-3)/2) \)
\( = 2ln 2 - 3 \)
