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

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

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

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

- \frac{x ^{2}}{2} + \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + C

Solution steps

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

Multiply by the conjugate of

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

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

Take the constant out:

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

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

Simplify

= - \int x - 1 + x^{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

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

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

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

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

Simplify

= - \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

= - \frac{x ^{2}}{2} + \frac{1}{2} x 1 + x^{2} + \frac{1}{2} ln x + 1 + x^{2} + C

\frac{1}{s ^{3}}

x^{2} y^{^{'}} - 2 x y = 3 y^{4}, y (1) = \frac{1}{2}

\int \frac{1}{e ^{x} + e ^{2 x}} d x

\int \frac{tan ^{3} ( ln ( x ) )}{7 x} d x

8 e^{3 x} + e^{- x^{2}}

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