How do I find the integral #intarctan(4x)dx# ?

Answer 1

#I=x*tan^-1(4x)-1/4log|sqrt(1+16x^2)|+C#
#=x*tan^-1(4x)-1/8log|(1+16x^2)|+C#

#(1)I=inttan^-1(4x)dx# Let, #tan^-1(4x)=urArr4x=tanurArr4dx=sec^2udu##rArrdx=1/4sec^2udu# #I=intu*1/4sec^2udu=1/4intu*sec^2udu# Using Integration by Parts, #I=1/4[u*intsec^2udu-int(d/(du)(u)*intsec^2udu)du]=1/4[u*tanu-int1*tanudu]##=1/4[u*tanu-log|secu|]+C##=1/4[tan^-1(4x)*(4x)-log|sqrt(1+tan^2u|]+C##=x*tan^-1(4x)-1/4log|sqrt(1+16x^2)|+C# Second Method: #(2)I=int1*tan^-1(4x)dx##=tan^-1(4x)*x-int(1/(1+16x^2)*4)xdx# #=x*tan^-1(4x)-1/8int(32x)/(1+16x^2)dx# #=x*tan^-1(4x)-1/8log|1+16x^2|+C#
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Answer 2

To find the integral of ( \int \arctan(4x) , dx ), you can use integration by parts. Let ( u = \arctan(4x) ) and ( dv = dx ). Then, ( du = \frac{1}{1+(4x)^2} , dx ) and ( v = x ).

Now, apply the integration by parts formula: [ \int u , dv = uv - \int v , du ]

Substitute the values of ( u ), ( dv ), ( du ), and ( v ) into the formula: [ \int \arctan(4x) , dx = x \arctan(4x) - \int \frac{x}{1+(4x)^2} , dx ]

The integral on the right side can be solved by substituting ( u = 1 + (4x)^2 ) and ( du = 8x , dx ): [ = x \arctan(4x) - \frac{1}{8} \int \frac{1}{u} , du ]

[ = x \arctan(4x) - \frac{1}{8} \ln|u| + C ]

Finally, substitute back ( u = 1 + (4x)^2 ) and simplify: [ = x \arctan(4x) - \frac{1}{8} \ln(1 + (4x)^2) + C ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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