(1+x) (1+x²) \, dx = ≤ft( x + \;{x²}{2} + \;1/2 ) (1+x²) + g(x) + C, then g(x)=

Correct Answer is : −2x− x22+2tan⁡−1x-2x - \;{x²}{2} + 2 ^{-1} x−2x−2x2+2tan−1x

2 ^{-1} x + \;{x²}{2} + \;{x³}{3}

Correct Answer is : −2x− x22+2tan⁡−1x-2x - \;{x²}{2} + 2 ^{-1} x−2x−2x2+2tan−1x

Test Index

AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 73 of 160

Marks: +1, -0

\int (1+x) \log (1+x^{2}) \, dx = \left( x + \;\frac{x^{2}}{2} + \;\frac{1}{2} \right) \log (1+x^{2}) + g(x) + C

g(x)=

Solution: 👈: Video Solution

Let

I = \int (1+x) \log (1+x^{2}) \, dx

Using by parts

\log (1+x^{2}) \left[ \frac{x^{2}}{2} + x \right] - \int \frac{2x}{1+x^{2}} \times \left( \frac{x^{2}}{2} + x \right) \, dx

= \left( \frac{x^{2}}{2} + x \right) \log (1+x^{2}) - \int \frac{x^{2} (x+2)}{1+x^{2}} \, dx

= \left( \frac{x^{2}}{2} + x \right) \log (1+x^{2}) - \int \frac{x^{3} + 2x^{2}}{1+x^{2}} \, dx

= \left( \frac{x^{2}}{2} + x \right) \log (1+x^{2}) - \int (x+2) - \frac{x+2}{1+x^{2}} \, dx

= \left( \frac{x^{2}}{2} + x \right) \log (1+x^{2}) - \left( \frac{x^{2}}{2} + 2x \right) + \frac{1}{2} \int \frac{2x}{1+x^{2}} \, dx

+ 2 \int \frac{1}{1+x^{2}} \, dx

= \left( \frac{x^{2}}{2} + x \right) \log (1+x^{2}) - \frac{x^{2}}{2} - 2x +

\frac{1}{2} \log (1+x^{2})

+ 2 \tan^{-1} x

= \left( \frac{x^{2}}{2} + x + \frac{1}{2} \right) \log (1+x^{2}) - \frac{x^{2}}{2} - 2x + 2 \tan^{-1} x + C

\text{Compare I with } \left( x + \frac{x^{2}}{2} + \frac{1}{2} \right)

\log (1+x^{2}) + g(x) + C, \text{ we get }

g(x) = -2x - \frac{x^{2}}{2} + 2 \tan^{-1} x

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