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Question Number 18024 by mondodotto@gmail.com last updated on 13/Jul/17

Answered by sma3l2996 last updated on 14/Jul/17

$$I=\int \frac{5x+3}{\sqrt{x^2+4x+10}}dx=\frac{5}{2}\int \frac{2x}{\sqrt{x^2+4x+10}}dx+\int \frac{3dx}{\sqrt{x^2+4x+10}}$$$$=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^2+4x+10}}+\int \frac{3-10}{\sqrt{{(x+2)}^2+6}}dx$$$$=5\sqrt{x^2+4x+10}-\int \frac{7}{\sqrt{6}\sqrt{{\left(\frac{x+2}{\sqrt{6}}\right)}^2+1}}dx+c$$$$lett=\frac{x+2}{\sqrt{6}}\Rightarrow dt=\frac{dx}{\sqrt{6}}$$$$\frac{7}{\sqrt{6}}\int \frac{dx}{\sqrt{{\left(\frac{x+2}{\sqrt{6}}\right)}^2+1}}=7\int \frac{dt}{\sqrt{t^2+1}}=7ln(t+\sqrt{t^2+1})+a$$$$I=5\sqrt{x^2+4x+10}-7ln(\frac{x+2}{\sqrt{6}}+\sqrt{{\left(\frac{x+2}{\sqrt{6}}\right)}^2+1})+C_1$$$$=5\sqrt{x^2+4x+10}-7ln(x+2+\sqrt{x^2+4x+10})-7ln\sqrt{6}+C_1$$$$I=5\sqrt{x^2+4x+10}-7ln(x+2+\sqrt{x^2+4x+10})+C$$

Commented by mondodotto@gmail.com last updated on 14/Jul/17

$$\mathrm{i}\mathrm{preciate}\mathrm{sir}\mathrm{thanx}$$

Answered by Abbas-Nahi last updated on 14/Jul/17

$$\int \frac{5x+3}{\sqrt{{(x+2)}^2+6}}dx$$$$letu=x+2du=dx$$$$\int \frac{5(u-2)+3}{\sqrt{u^2+6}}du=\int \frac{5u-7}{\sqrt{u^2+6}}du$$$$5\int \frac{u}{\sqrt{u^2+6}}du-7\int \frac{1}{\sqrt{u^2+6}}du$$$$\frac{5}{2}\int \frac{2u}{\sqrt{u^2+6}}du-\frac{7}{\sqrt{6}}\int \frac{1}{\sqrt{1+{\left(\frac{u}{\sqrt{6}}\right)}^2}}du$$$$thencompletethesolution....$$

Commented by mondodotto@gmail.com last updated on 14/Jul/17

$$\mathrm{waoohh}\mathrm{nice}1\mathrm{thanx}$$

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