For x0, evaluate the integral: int left( fra | vedclass

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(D) Let $I = \int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$. Combine the first and t

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$\sqrt{x^2+x+1}+C$

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(D) Let $I = \int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$. Combine the first and third terms: $I = \int \left( \frac{1+x+x^2-1}{(1+x) \sqrt{1+x+x^2}} \right) dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$. Simplify the numerator: $I = \int \frac{x(1+x)}{(1+x) \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{x}{\sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$. Multiply and divide the first integral by $2$: $I = \int \frac{2x}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{2x+1-1}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx$. Split the integral: $I = \int \frac{2x+1}{2 \sqrt{1+x+x^2}} dx - \int \frac{1}{2 \sqrt{1+x+x^2}} dx + \int \frac{1}{2 \sqrt{1+x+x^2}} dx = \int \frac{2x+1}{2 \sqrt{1+x+x^2}} dx$. Let $t = x^2+x+1$, then $dt = (2x+1) dx$. $I = \frac{1}{2} \int \frac{1}{\sqrt{t}} dt = \sqrt{t} + C = \sqrt{x^2+x+1} + C$.

For $x>0$, evaluate the integral: $\int \left( \frac{\sqrt{1+x+x^2}}{1+x} + \frac{1}{2 \sqrt{1+x+x^2}} - \frac{1}{(1+x) \sqrt{1+x+x^2}} \right) dx$.

Similar Questions

Observe the following statements :
$A: \int \left(\frac{x^2-1}{x^2}\right) e^{\frac{x^2+1}{x}} d x = e^{\frac{x^2+1}{x}} + c$
$R: \int f^{\prime}(x) e^{f(x)} d x = f(x) + c$
Then which of the following is true?

If $\int \frac{\sqrt{1-x^4}}{x^7} d x=f(x)\left\{\sqrt{1-x^4}\right\}^n+C$,then $(f(x))^n$ is equal to

$\begin{aligned} & \text{If } 5(f(x))^2 = x f(x) + 30 \text{ and } \\ & \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx \\ & = \frac{A}{B x^3 + D f(x)} + C, \text{ then } A + B + D = \end{aligned}$

$\int\left(1+x-x^{-1}\right) e^{x+x^{-1}} d x$ is equal to :

$\int \frac{2 \sin x - 3 \cos x}{4 \cos x - 3 \sin x} dx = $

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