Question

Find \(\int \frac{(x+3)}{2x^2+6x+7} dx\).

(a) \(\frac{1}{4} log⁡(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}}\right )+C\)

(b) \(\frac{1}{4} log⁡(2x^2+6x+7) – \frac{3}{4} (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}} )+C\)

(c) \(log⁡(2x^2+6x+7) + \left (tan^{-1}⁡\frac{2x+3}{2√2}\right )+C\)

(d) –\(log⁡(2x^2+6x+7) – \frac{3}{4} \left (\frac{1}{√2} tan^{-1}⁡\frac{2x+3}{2√2}\right )+C\)

The question was asked in an online quiz.

The question is from Integrals of Some Particular Functions in portion Integrals of Mathematics – Class 12

Answer 1

Right answer is (a) \(\frac{1}{4} log⁡(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}}\right )+C\)

Explanation: We can express

x+3=A \(\frac{d}{dx}\) (2x^2+6x+7)+B

x+3=A(4x+6)+B

x+3=4Ax+(6A+B)

Comparing the coefficients, we get

4A=1 ⇒A=1/4

6A+B=3 ⇒B=3/2

\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx+\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx\)

Let 2x^2+6x+7=t

(4x+6)dx=dt

\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} \int \frac{dt}{t}=\frac{1}{4} log⁡t\)

Replacing t with (2x^2+6x+7)

\(\frac{1}{4} \int \frac{4x+6}{2x^2+6x+7} dx=\frac{1}{4} log⁡(2x^2+6x+7)\)

\(\frac{3}{2} \int \frac{1}{2x^2+6x+7} dx=\frac{3}{2} \int \frac{1}{2(x^2+3x+\frac{7}{2})} dx=\frac{3}{4} \int \frac{1}{(x+\frac{3}{2})^2+2} dx\)

=\(\frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2\sqrt{2}} \right )\)

∴\(\int \frac{x+3}{2x^2+6x+7} dx=\frac{1}{4} log⁡(2x^2+6x+7) + \frac{3}{4} \left (\frac{1}{\sqrt{2}} tan^{-1}⁡\frac{2x+3}{2√2} \right )+C\)