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

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

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integral of (ln(x+sqrt(1+x^2)))'

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

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

Solution steps

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

Apply Integration By Parts

: x (ln (x + 1 + x^{2}))^{'} - \int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

= x (ln (x + 1 + x^{2}))^{^{'}} - \int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x

\int - \frac{x ^{2}}{( 1 + x ^{2} ) ^{\frac{3}{2}}} d x = - ln x + 1 + x^{2} + \frac{x}{1 + x ^{2}}

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

Simplify

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

Add a constant to the solution

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

\frac{\partial}{\partial u} (u^{\frac{1}{3}} \cdot v^{\frac{2}{3}})

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

4 x^{3} - 5 7 x^{6} + π x

e^{- 4 t} cos (3 t)

\int x e^{3 x} d x

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