The value of the integral $\int_{-1}^{2} \log _{e}\left(x+\sqrt{x^{2}+1}\right) d x$ is :

(B) $\sqrt{2}-\sqrt{5}+\log _{\mathrm{e}}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

(A) $\sqrt{5}-\sqrt{2}+\log _{e}\left(\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right)$

(C) $\sqrt{5}-\sqrt{2}+\log _{e}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$

(D) $\sqrt{2}-\sqrt{5}+\log _{e}\left(\frac{7+4 \sqrt{5}}{1+\sqrt{2}}\right)$

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Indefinite Integrals

The value of the integral

\int_{- 1}^{2} lo g_{e} (x + x^{2} + 1) d x

is :

Answer: (B)

Using

\int ln (x + x^{2} + 1) d x = x ln (x + x^{2} + 1) - x^{2} + 1 + C

(by parts, since

\frac{d}{d x} ln (x + x^{2} + 1) = \frac{1}{x ^{2} + 1}

I = [x ln (x + x^{2} + 1) - x^{2} + 1]_{- 1}^{2}

x = 2

2 ln (2 + 5) - 5

.

At

x = - 1

- ln (2 - 1) - 2 = ln (\frac{1}{2 - 1}) - 2 = ln (2 + 1) - 2

I = 2 ln (2 + 5) - 5 - ln (2 + 1) + 2 = ln \frac{( 2 + 5 ) ^{2}}{2 + 1} + 2 - 5

= 2 - 5 + ln \frac{9 + 4 5}{1 + 2}

JEE Main · 2024 · 09 Apr 2024 · Shift 2 · Single correct option