Question

Integrate the following: $\displaystyle \int \frac{x^2}{x^2+x+3}\, dx$​

Answer 1

Question :

$\displaystyle \sf \dagger\ \; \red{ \int \dfrac{x^2}{x^2+x+3}\ dx}$

Solution :

$\displaystyle \sf \int \dfrac{x^2}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{x^2+x+3-x-3}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf \int \dfrac{x^2+x+3}{x^2+x+3}\ dx- \int \dfrac{(x+3)}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf \int dx- \dfrac{2}{2} \int \dfrac{(x+3)}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \dfrac{2x+6}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \dfrac{2x+1+5}{x^2+x+3}\ dx$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \left( \dfrac{2x+1}{x^2+x+3} + \dfrac{5}{x^2+x+3} \right)\ dx$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \dfrac{2x+1}{x^2+x+3}\ dx - \dfrac{5}{2} \int \dfrac{1}{x^2+x+3} \right)\ dx$

Use sub. method for 2nd part ,

➠ u = x² + x + 3

➠ du = ( 2x + 1 ) dx

Use completing square method for 3rd part ,

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \dfrac{du}{u} - \dfrac{5}{2} \int \dfrac{1}{(x+\frac{1}{2})^2+\frac{11}{4}} \right)\ dx$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2} \int \dfrac{du}{u} - \dfrac{5}{2} \int \dfrac{1}{(\frac{\sqrt{11}}{2})^2+(x+\frac{1}{2})^2} \right)\ dx$

$\displaystyle \bullet\ \; \sf \orange{\int \dfrac{dx}{x} = \ln x}$

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2}\ \ln u - \dfrac{5}{2} \int \dfrac{1}{(\frac{\sqrt{11}}{2})^2+(x+\frac{1}{2})^2} \right)\ dx$

For 3rd part , use sub. method again ,

➠ v = $\sf x + \dfrac{1}{2}$

➠ dv = dx

$\longrightarrow \displaystyle \sf x- \dfrac{1}{2}\ \ln u - \dfrac{5}{2} \int \dfrac{dv}{(\frac{\sqrt{11}}{2})^2+v^2} \right)$

$\displaystyle \bullet\ \; \sf \green{\int \dfrac{dx}{a^2+x^2} = tan^{-1} \left(\dfrac{x}{a} \right)}$

$\\ \longrightarrow \displaystyle \sf x- \dfrac{1}{2} \ln u - \dfrac{5}{2} . \dfrac{1}{\frac{\sqrt{11}}{2}}\ tan^{-1} \left( \dfrac{v}{\frac{\sqrt{11}}{2}} \right)+c$

$\\ \longrightarrow \displaystyle \sf x- \dfrac{1}{2} \ln\ (x^2+x+3) - \dfrac{5}{2} . \dfrac{1}{\frac{\sqrt{11}}{2}}\ tan^{-1} \left( \dfrac{x+\frac{1}{2}}{\frac{\sqrt{11}}{2}} \right)+c$

$\longrightarrow \displaystyle \sf \pink{x- \dfrac{1}{2} \ln\ (x^2+x+3) - \dfrac{5}{\sqrt{11}}\ tan^{-1} \left( \dfrac{2x+1}{\sqrt{11}} \right)+c}$

★ ═════════════════════ ★

$\displaystyle \sf \blue{\dagger\ \; \int \dfrac{x^2}{x^2+x+3}\ dx = x- \dfrac{1}{2} \ln\ (x^2+x+3) - \dfrac{5}{\sqrt{11}}\ tan^{-1} \left( \dfrac{2x+1}{\sqrt{11}} \right)+c }$

Answer 2

Answer:

Refer the attachment

• hope it helps