Find The Antiderivative of more complex expressions.Need help finding the antiderivative of this one in...

gaiaecologicaq2

gaiaecologicaq2

Answered

2022-07-09

Find The Antiderivative of more complex expressions.
Need help finding the antiderivative of this one in particular d x 3 x ² + 1 .
Would you kindly also give some tips on how to find the antiderivative of any expressions. I am having problems switching from differential calculus to integral calculus as finding the antiderivative is not as straight forward as finding the differential of an expression.

Answer & Explanation

pampatsha

pampatsha

Expert

2022-07-10Added 15 answers

Step 1
Generalize this. Start with the integral
I = d x a 2 + ( b x ) 2
Let z = b x a so that d z = b a d x. Thus
I = a b d z a 2 + a 2 z 2 = 1 a b d z 1 + z 2
Step 2
The integral is now easily seen to be arctan z + C. Thus
d x a 2 + ( b x ) 2 = 1 a b arctan ( b x a ) + C
Now set a = 1 and b = 3 .
Holetaug

Holetaug

Expert

2022-07-11Added 8 answers

Step 1
You need to know the formula
1 x 2 + a 2 d x = 1 a arctan ( x a ) + c
So your integral becomes:
1 3 x 2 + 1 d x = 1 3 1 x 2 + 1 3 d x = 1 3 1 x 2 + ( 1 3 ) 2 d x = 1 3 × 1 1 / 3 arctan ( x 1 / 3 ) + c = 1 3 arctan 3 x + c
Step 2
ANOTHER METHOD: By Substitution
Let x = 1 3 tan θ . So d x = 1 3 sec 2 θ   d θ . Also 3 x 2 + 1 = 3 ( 1 3 tan 2 θ ) + 1 = tan 2 θ + 1 = sec 2 θ.
Hence your integral becomes
1 3 x 2 + 1 d x = 1 sec 2 θ ( 1 3 sec 2 θ   d θ ) = 1 3 d θ = 1 3 θ + c                     ( i )
Now since x = 1 3 tan θ , so θ = arctan 3 x. Putting this in one we have the required answer as 1 3 x 2 + 1 d x = 1 3 arctan 3 x + c.

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