Mathematics

# Solve:$\displaystyle\int { \sqrt { ax+b } } dx$

##### SOLUTION
$\int \sqrt{ax+b}.dx$
Let $ax+b = t$
diff w.r.t 'x'
$a(1)+0=\frac{dt}{dx}$
$dx=\frac{dt}{a}$
now
$\int \sqrt{t}.\frac{dt}{a}$
= $\frac{1}{a}\int \sqrt{t}.dt$
= $\frac{1}{a}\int t^{1/2}.dt$
= $\frac{1}{a}[\frac{t^{3/2}}{3/2}+C]$
= $\frac{1}{a}.[\frac{2}{3}t^{3/2}+C]$
= $\frac{2}{3a}t^{3/2}+C$
= $\frac{2}{3a}(ax+b)^{3/2}+C$

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Subjective Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 105

#### Realted Questions

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