What is the integral of (1+sqrt(x^2+1))/((x^2+1)^{3/2)} ?

The integral of (1+sqrt(x^2+1))/((x^2+1)^{3/2)} is x/(sqrt(1+x^2))+arctan(x)+C

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integral of (1+sqrt(x^2+1))/((x^2+1)^{3/2)}

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

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

Solution steps

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

Expand

\frac{1 + x ^{2} + 1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} : \frac{1}{( x ^{2} + 1 ) ^{\frac{3}{2}}} + \frac{1}{x ^{2} + 1}

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

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

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

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

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

Apply radical rule:

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

Add a constant to the solution

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

What is the integral of (1+sqrt(x^2+1))/((x^2+1)^{3/2)} ?
The integral of (1+sqrt(x^2+1))/((x^2+1)^{3/2)} is x/(sqrt(1+x^2))+arctan(x)+C