dx-p-qx-r-

Question Number 94161 by i jagooll last updated on 17/May/20

\int \frac{dx}{p + \sqrt{qx + r}}

Commented by mathmax by abdo last updated on 17/May/20

I = \int \frac{d x}{p + \sqrt{q x + r}} w e u s e t h e c h a n g e m e m t \sqrt{q x + r} = t \Rightarrow

q x + r = t^{2} \Rightarrow q x = t^{2} - r \Rightarrow d x = \frac{2 t}{q} d t \Rightarrow

I = \frac{2}{q} \int \frac{t d t}{p + t} = \frac{2}{q} \int \frac{p + t - p}{p + t} d t = \frac{2}{q} t - \frac{2 p}{q} \int \frac{d t}{t + p}

= \frac{2}{q} t - \frac{2 p}{q} l n ∣ t + p ∣ + C

= \frac{2 \sqrt{q x + r}}{q} - \frac{2 p}{q} l n ∣ p + \sqrt{q x + r} ∣ + C (w e s u p p o s e q \neq 0)

i f q = 0 \Rightarrow I = \frac{x}{p + \sqrt{r}}

Commented by hknkrc46 last updated on 17/May/20

\int \frac{dx}{p + \sqrt{qx + r}} {\begin{cases} p + \sqrt{qx + r} = t \\ x = \frac{{(t - p)}^{2} - r}{q} \\ dx = \frac{2 (t - p)}{q} dt \end{cases}

= \frac{2}{q} \int \frac{t - p}{t} dt = \frac{2}{q} \int (1 - \frac{p}{t}) dt

= \frac{2 t}{q} - \frac{2 p}{q} \ln ∣ t ∣ + C_{1} = - \ln ∣ p + \sqrt{qx + r} ∣^{\frac{2 p}{q}} + \ln ∣ C ∣

{\ln ∣ C ∣= C_{1} + \frac{2 t}{q}}

= \ln ∣ \frac{C}{{(p + \sqrt{qx + r})}^{\frac{2 p}{q}}} ∣

Answered by john santu last updated on 17/May/20

set \sqrt{qx + r} = pt \Rightarrow qx + r = p^{2} t^{2}

dx = \frac{{2 p}^{2} t}{q} dt

I = \int \frac{{2 p}^{2} t dt}{q (p + pt)} = \frac{2 p}{q} \int \frac{t}{1 + t} dt

I = \frac{2 p}{q} \int \frac{t + 1 - 1}{t + 1} dt = \frac{2 p}{q} (t - \ln ∣ t + 1 ∣ + c

I = \frac{2 p}{q} t - \frac{2 p}{q} \ln ∣ t + 1 ∣ + c

I = \frac{2 p \sqrt{qx + r}}{pq} - \frac{2 p}{q} \ln ∣ \frac{p + \sqrt{qx + r}}{p} ∣ + c

I = \frac{2 \sqrt{qx + r}}{q} - \frac{2 p}{q} \ln ∣ \frac{p + \sqrt{qx + r}}{p} ∣ + c

Commented by i jagooll last updated on 17/May/20

thank you

Commented by peter frank last updated on 04/Jun/20

thank you