How can i solve this integral? ∫(25x√3+4x2dx

1 Answer
mason m
Sep 2, 2017

I got: 2√35ln∣∣∣3x2+42√3x+√3x2−2√3x∣∣∣+C

Explanation:

I=∫25x√3+4x2dx

Use the substitution x=2√3cotθ. This implies that dx=−2√3cotθcscθdθ.

I=25∫√32tanθ ⎷3+4(√32tanθ)2(−2√3cotθcscθdθ)

Canceling √32tanθ(2√3cotθ)=1:

I=−25∫cscθ√3+3tan2θdθ

Factoring √3, the radical becomes √1+tan2θ=secθ:

I=−2√35∫dθsinθcosθ

Note that sinθcosθ=12sin2θ:

I=−4√35∫dθsin2θ

I=−4√35∫csc2θdθ

The integral of cosecant is fairly standard. Look it up here if you aren't sure.

Note that I'll use the form ∫cscx=−ln|cscx+cotx|, which is equivalent to ln|cscx−cotx|. I'm choosing the first one so the minus sign will cancel with the minus sign already outside the integral.

You'll also need to undo the 2θ by multiplying the integral by 1/2.

I=2√35ln|csc2θ+cot2θ|+C

Use csc2θ=1sin2θ=12sinθcosθ and tan2θ=2tanθ1−tan2θ:

I=2√35ln∣∣∣12sinθcosθ+1−tan2θ2tanθ∣∣∣+C

Before proceeding we can factor 2 from both these denominators and remove −2√35ln2 from the logarithm using log(A/B)=logA−logB. This constant is absorbed into C.

I=2√35ln|cscθsecθ+cotθ−tanθ|+C

Our original substitution was cotθ=√3x2. Thus:

  • tanθ=1cotθ=2√3x
  • cscθ=√cot2θ+1=√3x2+42
  • secθ=√tan2θ+1=√3x2+4√3x

So:

I=2√35ln∣∣∣3x2+42√3x+√3x2−2√3x∣∣∣+C

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