Question Number 158768 by EbrimaDanjo last updated on 08/Nov/21
$$ \\ $$$$\mathrm{evaluate} \\ $$$$\int\mathrm{2x}\sqrt{\mathrm{4x}−\mathrm{5}}\:\mathrm{dx} \\ $$
Answered by MJS_new last updated on 08/Nov/21
![∫2x(√(4x−5))dx=? without substitution: (d/dx)[(ax+b)(4x−5)^(3/2) ]=(10ax−5a+6b)(√(4x−5)) ⇒ 10ax−5a+6b=2x ⇒ a=(1/5)∧b=(1/6) ⇒ ∫2x(√(4x−5))dx=((x/5)+(1/6))(4x−5)^(3/2) +C](Q158769.png)
$$\int\mathrm{2}{x}\sqrt{\mathrm{4}{x}−\mathrm{5}}{dx}=? \\ $$$$\mathrm{without}\:\mathrm{substitution}: \\ $$$$\frac{{d}}{{dx}}\left[\left({ax}+{b}\right)\left(\mathrm{4}{x}−\mathrm{5}\right)^{\mathrm{3}/\mathrm{2}} \right]=\left(\mathrm{10}{ax}−\mathrm{5}{a}+\mathrm{6}{b}\right)\sqrt{\mathrm{4}{x}−\mathrm{5}} \\ $$$$\Rightarrow \\ $$$$\mathrm{10}{ax}−\mathrm{5}{a}+\mathrm{6}{b}=\mathrm{2}{x} \\ $$$$\Rightarrow \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{5}}\wedge{b}=\frac{\mathrm{1}}{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$$\int\mathrm{2}{x}\sqrt{\mathrm{4}{x}−\mathrm{5}}{dx}=\left(\frac{{x}}{\mathrm{5}}+\frac{\mathrm{1}}{\mathrm{6}}\right)\left(\mathrm{4}{x}−\mathrm{5}\right)^{\mathrm{3}/\mathrm{2}} +{C} \\ $$
Commented by EbrimaDanjo last updated on 08/Nov/21
$$\mathrm{thanks}\:\mathrm{sir} \\ $$