İntegral

[ArtOfMathSolving]

$$\int^{0}_{-1} \frac{x^2+2x}{\ln(x+1)} \,dx$$ ifadesinin değerini hesaplayın.

[stuart clark]

$\displaystyle I = \int_{-1}^{0}\frac{x^2+2x}{\ln(x+1)}dx = \int_{-1}^{0}\frac{(x+1)^1-1}{\ln(x+1)}dx$

Now Put $x+1=t$ and $dx = dt$ and changing Limits, We get

$\displaystyle I = \int_{0}^{1}\frac{t^2-1}{\ln t}dt = \int_{0}^{1}\int_{0}^{1}x^{2t}dtdx = \int_{0}^{1}\int_{0}^{1}x^{2t}dxdt$

So we get $\displaystyle I = \int_{0}^{1}\left[\frac{x^{2t+1}}{2t+1}\right]_{0}^{1}dx = \int_{0}^{1}\frac{1}{2t+1}dt = \left[\ln(2t+1)\right]_{0}^{1}$

So we get $\displaystyle I =  \int_{-1}^{0}\frac{x^2+2x}{\ln(x+1)}dx  = \ln(3)$