What is the integral of 1/(sqrt(x^2+2))x^2+1) ?

The integral of 1/(sqrt(x^2+2))x^2+1) is 1/2 dxsqrt(x^2+2)+C

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integral of 1/(sqrt(x^2+2))x^2+1)

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

\frac{1}{2} d x x^{2} + 2 + C

Solution steps

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

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

Simplify

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

= d \cdot \int \frac{x ^{2} + 1}{x ^{2} + 2} d x

Expand

\frac{x ^{2} + 1}{x ^{2} + 2} : \frac{x ^{2}}{x ^{2} + 2} + \frac{1}{x ^{2} + 2}

= d \cdot \int \frac{x ^{2}}{x ^{2} + 2} + \frac{1}{x ^{2} + 2} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

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

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

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

Simplify

= \frac{1}{2} d x x^{2} + 2

Add a constant to the solution

= \frac{1}{2} d x x^{2} + 2 + C

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What is the integral of 1/(sqrt(x^2+2))x^2+1) ?
The integral of 1/(sqrt(x^2+2))x^2+1) is 1/2 dxsqrt(x^2+2)+C