What is the integral of (rsqrt(ln(r^2+1)))/(r^2+1) ?

The integral of (rsqrt(ln(r^2+1)))/(r^2+1) is 1/3 ln^{3/2}(r^2+1)+C

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integral of (rsqrt(ln(r^2+1)))/(r^2+1)

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

\frac{1}{3} ln^{\frac{3}{2}} (r^{2} + 1) + C

Solution steps

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

Apply u-substitution

: \int u^{2} d u

= \int u^{2} d u

Apply the Power Rule

: \frac{u ^{3}}{3}

= \frac{u ^{3}}{3}

Substitute back

u = ln (r^{2} + 1)

= \frac{( ln ( r ^{2} + 1 ) ) ^{3}}{3}

Simplify

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

= \frac{1}{3} ln^{\frac{3}{2}} (r^{2} + 1)

Add a constant to the solution

= \frac{1}{3} ln^{\frac{3}{2}} (r^{2} + 1) + C

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What is the integral of (rsqrt(ln(r^2+1)))/(r^2+1) ?
The integral of (rsqrt(ln(r^2+1)))/(r^2+1) is 1/3 ln^{3/2}(r^2+1)+C