Ohh! my Integration

Calculus Level 2

Evaluate ∫ 3 x 2 + 2 x + 5 x 2 + 4 x + 6 d x . \large \int \dfrac{3x^2 + 2x+5}{x^2 + 4x+6} \, dx .

Clarification : C C in arbitrary constant of integration.

3 x − 5 ln ⁡ ∣ x 2 + 4 x − 6 ∣ + 7 2 T A N X − 1 x + 2 2 + C 3x-5 \ln \mid x^2 + 4x - 6 \mid + \dfrac{7}{\sqrt{2}}\ce{TAN^{-1}}\dfrac{x+2}{\sqrt{2}}+C 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 2 T A N X − 1 x + 2 2 + C 3x-5 \ln \mid x^2 + 4x + 6 \mid + \dfrac{7}{\sqrt{2}}\ce{TAN^{-1}}\dfrac{x+2}{\sqrt{2}}+C 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 2 T A N X − 1 x − 2 2 + C 3x-5 \ln \mid x^2 + 4x + 6 \mid + \dfrac{7}{\sqrt{2}}\ce{TAN^{-1}}\dfrac{x-2}{\sqrt{2}}+C

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1 solution

Those who are beginners and interested of calculus

Steps for ∫ 3 x 2 + 2 x + 5 x 2 + 4 x + 6 d x \int \frac{3x^2+2x+5}{x^2+4x+6}\,dx

∫ 3 x 2 + 2 x + 5 x 2 + 4 x + 6 d x = ∫ ( 3 − 10 x + 13 x 2 + 4 x + 6 ) Use polynomial division = ∫ 3 d x − ∫ 10 x + 13 x 2 + 4 x + 6 d x u = a x 2 + b x + c ; d u = 2 x + 4 = 3 x − ∫ 10 x + 20 x 2 + 4 x + 6 d x + 7 ∫ 1 x 2 + 4 x + 6 d x = 3 x − 5 ln ⁡ ∣ u ∣ + 7 ∫ 1 x 2 + 4 x + 6 d x = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 ∫ 1 x 2 + 4 x + 6 a x 2 + b x + c = c + ( x a + b 2 a ) 2 − b 2 4 a = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 ∫ 1 ( x + 2 ) 2 + 2 = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 ∫ 1 u 2 + 2 d u u = x + 2 ; d u = 1 = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 ∫ 1 ( x + 2 ) 2 + 2 = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 2 2 ∫ 1 v 2 + 1 d v v = 2 2 u = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 2 2 T A N − 1 v Substitule back all = 3 x − 5 ln ⁡ ∣ x 2 + 4 x + 6 ∣ + 7 2 2 T A N − 1 ( 2 2 ( x + 2 ) ) + C □ \begin{aligned} \int \frac{3x^2+2x+5}{x^2+4x+6}\,dx &=\int \left(3-\frac{10x+13}{x^2+4x+6}\right) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{\text{Use polynomial division}} \\&= \int 3\,dx - \int \frac{10x+13}{x^2+4x+6}\,dx \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad{u=ax^2+bx+c; du=2x+4}\\&= 3x - \int \frac{10x+20}{x^2+4x+6}\,dx + 7 \int \frac{1}{x^2+4x+6}\,dx \\&=3x-5\ln \mid u \mid +7\int \frac{1}{x^2+4x+6}\,dx \\&= 3x-5 \ln \mid x^2+4x+6 \mid +7 \int \frac{1}{x^2+4x+6} \quad\quad\quad\quad{ax^2+bx+c = c+\left(x\sqrt{a}+\frac{b}{2\sqrt{a}}\right)^2-\frac{b^2}{4a}}\\&= 3x-5 \ln \mid x^2+4x+6 \mid +7 \int \frac{1}{(x+2)^2+2} \\&= 3x-5 \ln \mid x^2+4x+6 \mid +7 \int \frac{1}{u^2+2}\,du \quad\quad\quad\quad\quad\quad\quad{u=x+2; du =1} \\&=3x-5 \ln \mid x^2+4x+6 \mid +7 \int \frac{1}{(x+2)^2+2} \\&= 3x-5 \ln \mid x^2+4x+6 \mid +\frac{7\sqrt{2}}{2} \int \frac{1}{v^2+1}\,dv \quad\quad\quad\quad\quad\quad\quad\quad\quad{v=\frac{\sqrt{2}}{2}u}\\&= 3x-5 \ln \mid x^2+4x+6 \mid +\frac{7\sqrt{2}}{2} \ce{TAN}^{-1}v \quad\quad\quad\quad\quad\quad{\text{Substitule back all}} \\&=3x-5 \ln \mid x^2+4x+6 \mid +\frac{7\sqrt{2}}{2}\ce{TAN}^{-1}\left( \frac{\sqrt{2}}{2}(x+2)\right)+\ce{C}\space \square \end{aligned}

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