Evaluate the integrals. ∫(tan−1x)5(1+x2)dx

Question

Evaluate the integrals.  ∫(tan−1x)5(1+x2)dx

Lakisha Archer · Accepted Answer

Step 1
We have to evaluate the integral:

\int \frac{{(\tan^{- 1} x)}^{5}}{(1 + x^{2})} d x

This integral will be solved by substitution method since derivative of one function is present in the integral.
Assuming,

t = \tan^{- 1} x

Differentiating with respect to x, we get

\frac{d t}{d x} = \frac{d (\tan^{- 1} x)}{d x}

= \frac{1}{1 + x^{2}}

d t = \frac{1}{1 + x^{2}} d x

Step 2
Substituting above values in the integral,

\int \frac{{(\tan^{- 1} x)}^{5}}{(1 + x^{2})} d x = \int t^{5} d t

= \frac{t^{5 + 1}}{5 + 1} + C

(since

\int x^{n} d x = \frac{x^{n + 1}}{n + 1} + C

)

= \frac{t^{6}}{6} + C

= \frac{1}{6} {(\tan^{- 1} x)}^{6} + C

Where, C is an arbitrary constant.
We have substituted value of t for last step.
Hence, value of integral is

\frac{1}{6} {(\tan^{- 1} x)}^{6} + C

.

David Clayton · Accepted Answer

∫(tan−1x)5(1+x2)dx=∫arctan5(x)x2+1dx  ∫arctan5(x)x2+1dx  Substitution u=arctan⁡(x)⇒dudx=1x2+1⇒dx=(x2+1)du:  =∫u5du  Integral of a power function:  ∫undu=un+1n+1 at n=5:  =u66  Reverse replacement u=arctan⁡(x):  =arctan6(x)6  ∫arctan5(x)x2+1dx  =arctan6(x)6+C

karton · Accepted Answer

∫(tan−1⁡x)5(1+x2)dx=∫arctan5⁡(x)x2+1dx Transform the expression ∫t5dt Use ∫xndx=xn+1n+1,n≠−1 to evaluate the integral t66 Substitute back t=arctan⁡(x) arctan⁡(x)66 Add C Solution arctan⁡(x)66+C

Evaluate the integrals. \int \frac{(\tan^{-1}x)^{5}}{(1+x^{2})}dx

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