Calculate the integral. ∫4x4x2+1dx

Question

Calculate the integral.  ∫4x4x2+1dx

Vaing1990 · Accepted Answer

Step 1  We have to evaluate ∫4x4x2+1dx  Let substitute t=4x2+1  Differentiate both sides with respect to x,  dtdx=ddx(4x2+1)  dtdx=8x  dt8x=dx  Step 2  Substituting the above value in the given integral,  ∫4x4x2+1dx=∫4xt⋅dt8x  =12∫1tdt  =12ln⁡(t)+C  =ln⁡(x)+C   (Using alog⁡x=log⁡xa)  Put back t=4x2+1, we get  ∫4x4x2+1dx=ln⁡(4x2+1)+C  Hence, required answer is ln⁡(4x2+1)+C

Muspee · Accepted Answer

Step 1: Use Constant Factor Rule: ∫cf(x)dx=c∫f(x)dx.  4∫x4x2+1dx  Step 2: Use Integration by Substitution on ∫x4x2+1dx.  Let u=4x2+1,du=8xdx, then xdx=18du  Step 3: Using u and du above, rewrite ∫x4x2+1dx.  ∫18udu  Step 4: Use Constant Factor Rule: ∫cf(x)dx=c∫f(x)dx.  18∫1udu  Step 5: The derivative of ln⁡x is 1x.  ln⁡u8  Step 6: Substitute u=4x2+1 back into the original integral.  ln⁡(4x2+1)8  Step 7: Rewrite the integral with the completed substitution.  ln⁡(4x2+1)2  Step 8: Add constant.  ln⁡(4x2+1)2+C

Calculate the integral. \int \frac{4x}{4x^{2}+1}dx

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