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

The integral of in(x+sqrt(x^2+1)) is in((x^2)/2+1/2 xsqrt(1+x^2)+1/2 ln|x+sqrt(1+x^2)|)+C

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

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

in (\frac{x ^{2}}{2} + \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2}) + C

Solution steps

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

Take the constant out:

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

= in \cdot \int x + x^{2} + 1 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

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

\int x d x = \frac{x ^{2}}{2}

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

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

Add a constant to the solution

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

\frac{\partial}{\partial y} (x e^{y - 1})

\frac{d x}{d t} = 1.5 - \frac{2 x}{100 + t}

\int x (x - 6)^{2} d x

\int 2 e^{- \frac{x}{5}} d x

\frac{\partial}{\partial y} (\frac{y}{( x + 2 ) ^{2}})

What is the integral of in(x+sqrt(x^2+1)) ?
The integral of in(x+sqrt(x^2+1)) is in((x^2)/2+1/2 xsqrt(1+x^2)+1/2 ln|x+sqrt(1+x^2)|)+C