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

Category: QuestionsFind \int \frac{(x+3)}{2x^2+6x+7} dx.
Editor">Editor Staff asked 11 months ago

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

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The question is from Integrals of Some Particular Functions in portion Integrals of Mathematics – Class 12
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Editor">Editor Staff answered 11 months ago

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