If f(x)=tan^(-1)x+lin sqrt(1+x)-ln sqrt(1-x) . the integral of -Turito

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If $f left parenthesis x right parenthesis equals t a n to the power of negative 1 end exponent invisible function application x plus l space l n invisible function application square root of 1 plus x end root minus l n square root of 1 minus x end root$ . The integral of $1 divided by 2 space f apostrophe left parenthesis x right parenthesis$ with respect to $x to the power of 4 end exponent$ is -

Maths-General

$- l n (1 - x^{4}) + c$
$l n (1 + x^{4}) + c$
$e^{- x^{4}} + c$
$e^{\sqrt{1 - x^{4}}} + c$

Answer:The correct answer is: $negative l n open parentheses 1 minus x to the power of 4 end exponent close parentheses plus c$

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$- l n (1 - x^{4}) + c$
$l n (1 + x^{4}) + c$
$e^{- x^{4}} + c$
$e^{\sqrt{1 - x^{4}}} + c$

Answer:The correct answer is: $negative l n open parentheses 1 minus x to the power of 4 end exponent close parentheses plus c$

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Related Questions to study

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

$not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fraction$ is

$not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fraction$ is

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

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If . The integral of with respect to is -

Answer:The correct answer is:

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Related Questions to study

Evaluate dx

Evaluate dx

is

is

=

=

Assertion : can never become positive.Reason : f(x) = sgn x is always a positive function.

Assertion : can never become positive.Reason : f(x) = sgn x is always a positive function.

dx equals:

dx equals:

If I = , then I equals:

If I = , then I equals:

The indefinite integral of is, for any arbitrary constant -

The indefinite integral of is, for any arbitrary constant -

If f = x + 2 then is equal to

If f = x + 2 then is equal to

Let , then is equal to:

Let , then is equal to:

If the remainders of the polynomial f(x) when divided by are 2,5 then the remainder of when divided by is

If the remainders of the polynomial f(x) when divided by are 2,5 then the remainder of when divided by is

If then B=

If then B=

Let a,b,c such that ,

Let a,b,c such that ,

If then

If then

The partial fractions of are

The partial fractions of are

The partial fractions of are

The partial fractions of are

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Answer:The correct answer is: $negative l n open parentheses 1 minus x to the power of 4 end exponent close parentheses plus c$

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

Evaluate $not stretchy integral fraction numerator 1 over denominator left parenthesis x to the power of 2 end exponent minus 4 right parenthesis square root of x plus 1 end root end fraction$ dx

$not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fraction$ is

$not stretchy integral fraction numerator e to the power of x end exponent d x over denominator cos invisible function application h x plus sin invisible function application h x end fraction$ is

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

$fraction numerator d x over denominator square root of 1 minus x to the power of 2 end exponent end root minus 1 end fraction$ =

Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

Assertion : $f left parenthesis x right parenthesis equals s g n invisible function application left parenthesis x minus vertical line x vertical line right parenthesis$ can never become positive.
Reason : f(x) = sgn x is always a positive function.

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

If I = $not stretchy integral fraction numerator d x over denominator x square root of 1 minus x to the power of 3 end exponent end root end fraction$ , then I equals:

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

The indefinite integral of $left parenthesis 12 s i n invisible function application x plus 5 c o s invisible function application x right parenthesis to the power of negative 1 end exponent$ is, for any arbitrary constant -

If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

If f $open parentheses fraction numerator 3 x minus 4 over denominator 3 x plus 4 end fraction close parentheses$ = x + 2 then $not stretchy integral f left parenthesis x right parenthesis d x$ is equal to

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

If the remainders of the polynomial f(x) when divided by $x minus 1 comma x minus 2$ are 2,5 then the remainder of $f left parenthesis x right parenthesis$ when divided by $left parenthesis x minus 1 right parenthesis left parenthesis x minus 2 right parenthesis$ is

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are

The partial fractions of $fraction numerator 6 x to the power of 4 end exponent plus 5 x to the power of 3 end exponent plus x to the power of 2 end exponent plus 5 x plus 2 over denominator 1 plus 5 x plus 6 x to the power of 2 end exponent end fraction$ are