\;{x+1}{{x²+x+1}}\;dx = [AP EAPCET 18 May 2024 Shift 1]

Correct Answer is : x2+x+1+12sin⁡−1(2x+13)+c{x²+x+1} + 1/2 ^{-1} ≤ft({2x+1}{{3}}) + cx2+x+1+21sin−1(32x+1)+c

1/2 {x²+x+1} + 1/2 ^{-1} ≤ft({x+2}{{3}}) + c

1/2 {x²+x+1} + {2}{{3}} ^{-1} ≤ft({2x+1}{{3}}) + c

{x²+x+1} + {2}{{3}} |x²+x+1| + c

{x²+x+1} + 1/2 ^{-1} ≤ft({2x+1}{{3}}) + c

Correct Answer is : x2+x+1+12sin⁡−1(2x+13)+c{x²+x+1} + 1/2 ^{-1} ≤ft({2x+1}{{3}}) + cx2+x+1+21sin−1(32x+1)+c

Test Index

Indefinite Integration

Section: Mathematics

Question : 88 of 89

Marks: +1, -0

\int\;\frac{x+1}{\sqrt{x^2+x+1}}\;dx =

$\frac{1}{2} \sqrt{x^2+x+1} + \frac{1}{2} \cosh^{-1} \left(\frac{x+2}{\sqrt{3}}\right) + c$
$\frac{1}{2} \sqrt{x^2+x+1} + \frac{2}{\sqrt{3}} \tan^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + c$
$\sqrt{x^2+x+1} + \frac{2}{\sqrt{3}} \log |x^2+x+1| + c$
$\sqrt{x^2+x+1} + \frac{1}{2} \sin^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + c$

Solution:

\text{ Let } t = \int\;\frac{x+1}{\sqrt{x^2+x+1}}\;dx

= \frac{1}{2} \int\;\frac{2x+1}{\sqrt{x^2+x+1}}\;dx + \frac{1}{2} \int\;\frac{1}{\sqrt{x^2+x+1}}\;dx

I = I_1 + I_2 (\text{ Let }) \;\; \ldots \;\text{ (i) }

Let

I_1 = \frac{1}{2} \int\;\frac{2x+1}{\sqrt{x^2+x+1}}\;dx

Let

x^2 + x + 1 = t^2

(2x+1)\,dr = 2\,dt

t_1 = \frac{1}{2} \int^{\;\frac{1}{t}} \underset{\overline{}}{x\,dt}

= \int dt = t = \sqrt{x^2+x+1}

I_2 = \frac{1}{2} \int\;\frac{1}{\sqrt{x^2+x+1}}\;dr

= \frac{1}{2} \int\;\frac{1}{\sqrt{\left(x+\frac{1}{2}\right)^2 + \frac{3}{4}}}\;dx

Let

x + \frac{1}{2} = \cos\,dx = d\,x

f_2 = \frac{1}{2} \int\;\frac{1}{\sqrt{N^2 + \frac{3}{4}}}

= \frac{1}{2} \sinh^{-1} \left(\frac{\pi}{\sqrt{3}}\right)

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

Hence.

I = \sqrt{x^2+x+1} + \frac{1}{2} \sinh^{-1} \left(\frac{2x+1}{\sqrt{3}}\right) + C

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