What is the integral of (1+4x^2)^{3/2} ?

The integral of (1+4x^2)^{3/2} is 1/4 x(4x^2+1)^{3/2}+3/8 xsqrt(4x^2+1)+3/16 ln|2x+sqrt(4x^2+1)|+C

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integral of (1+4x^2)^{3/2}

\int (1 + 4 x^{2})^{\frac{3}{2}} d x

\frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1 + C

Solution steps

\int (1 + 4 x^{2})^{\frac{3}{2}} d x

Apply Integration By Parts

: x (1 + 4 x^{2})^{\frac{3}{2}} - \int 12 x^{2} (4 x^{2} + 1)^{\frac{1}{2}} d x

= x (1 + 4 x^{2})^{\frac{3}{2}} - \int 12 x^{2} (4 x^{2} + 1)^{\frac{1}{2}} d x

\int 12 x^{2} (4 x^{2} + 1)^{\frac{1}{2}} d x = \frac{3}{16} (4 x (4 x^{2} + 1)^{\frac{3}{2}} - 2 x 4 x^{2} + 1 - ln 2 x + 4 x^{2} + 1)

= x (1 + 4 x^{2})^{\frac{3}{2}} - \frac{3}{16} (4 x (4 x^{2} + 1)^{\frac{3}{2}} - 2 x 4 x^{2} + 1 - ln 2 x + 4 x^{2} + 1)

Simplify

x (1 + 4 x^{2})^{\frac{3}{2}} - \frac{3}{16} (4 x (4 x^{2} + 1)^{\frac{3}{2}} - 2 x 4 x^{2} + 1 - ln 2 x + 4 x^{2} + 1) : \frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1

= \frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1

Add a constant to the solution

= \frac{1}{4} x (4 x^{2} + 1)^{\frac{3}{2}} + \frac{3}{8} x 4 x^{2} + 1 + \frac{3}{16} ln 2 x + 4 x^{2} + 1 + C

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What is the integral of (1+4x^2)^{3/2} ?
The integral of (1+4x^2)^{3/2} is 1/4 x(4x^2+1)^{3/2}+3/8 xsqrt(4x^2+1)+3/16 ln|2x+sqrt(4x^2+1)|+C