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
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