Question Number 97091 by Mathudent last updated on 06/Jun/20
$${solve}\:\int{x}^{{x}+\mathrm{1}} {dx}\:. \\ $$
Answered by Sourav mridha last updated on 06/Jun/20
![∫e^((x+1).ln(x)) dx =∫Σ_(n=0) ^∞ (((x+1)^n [ln(x)]^n )/(n!))dx =Σ_(n=0) ^∞ (1/(n!))[Σ_(r=0) ^n C_r ^n ∫x^(n−r) (ln(x))^n dx] let ln(x)=k... =Σ_(n=0) ^∞ (1/(n!))[Σ_(r=0) ^n C_r ^n ∫e^((n−r+1).k) k^n dk] =Σ_(n=o) ^∞ (1/(n!))[Σ_(r=0) ^n C_r ^n {Σ_(m=0) ^∞ (((n−r+1)^m )/(m!)).(k^((m+n+1)) /((m+n+1)))}]+g](Q97119.png)
$$\int\boldsymbol{{e}}^{\left(\boldsymbol{{x}}+\mathrm{1}\right).\boldsymbol{{l}}\mathrm{n}\left(\boldsymbol{{x}}\right)} \boldsymbol{{dx}} \\ $$$$=\int\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\boldsymbol{{x}}+\mathrm{1}\right)^{\boldsymbol{{n}}} \left[\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right]^{\boldsymbol{{n}}} }{\boldsymbol{{n}}!}\boldsymbol{{dx}} \\ $$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\left[\underset{\boldsymbol{{r}}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\overset{\boldsymbol{{n}}} {\boldsymbol{{C}}}_{\boldsymbol{{r}}} \int\boldsymbol{{x}}^{\boldsymbol{{n}}−\boldsymbol{{r}}} \left(\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)\right)^{\boldsymbol{{n}}} \boldsymbol{{dx}}\right] \\ $$$$\boldsymbol{{let}}\:\boldsymbol{{ln}}\left(\boldsymbol{{x}}\right)=\boldsymbol{{k}}... \\ $$$$=\underset{\boldsymbol{{n}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\left[\underset{\mathrm{r}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\overset{\boldsymbol{{n}}} {\boldsymbol{{C}}}_{\boldsymbol{{r}}} \int\boldsymbol{{e}}^{\left(\boldsymbol{{n}}−\boldsymbol{{r}}+\mathrm{1}\right).\boldsymbol{{k}}} \boldsymbol{{k}}^{\boldsymbol{{n}}} \boldsymbol{{dk}}\right] \\ $$$$=\underset{\boldsymbol{{n}}=\boldsymbol{{o}}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\boldsymbol{{n}}!}\left[\underset{\mathrm{r}=\mathrm{0}} {\overset{\boldsymbol{{n}}} {\sum}}\overset{\mathrm{n}} {\boldsymbol{{C}}}_{\boldsymbol{{r}}} \left\{\underset{\boldsymbol{{m}}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\boldsymbol{{n}}−\boldsymbol{{r}}+\mathrm{1}\right)^{\boldsymbol{{m}}} }{\boldsymbol{{m}}!}.\frac{\boldsymbol{{k}}^{\left(\boldsymbol{{m}}+\boldsymbol{{n}}+\mathrm{1}\right)} }{\left(\boldsymbol{{m}}+\boldsymbol{{n}}+\mathrm{1}\right)}\right\}\right]+\boldsymbol{{g}} \\ $$
Commented by Mathudent last updated on 07/Jun/20
$${thank}\:{you} \\ $$