If f(x)= 1x^1/4(1+x^1/4)dx,f(0)=-6, then f(1) is equal to :

Question

If $f(x)=\int \frac{1}{{x}^{1/4}(1+{x}^{1/4})}dx,f(0)=-6$, then $f(1)$ is equal to :

Accepted Answer

Substitution: Start by substituting $x={t}^{4}$ which implies $dx=4{t}^{3}\, dt$.

Integral Transformation:

$\int \frac{1}{{x}^{1/4}(1+{x}^{1/4})}\, dx\Rightarrow \int \frac{4{t}^{3}\, dt}{t(1+t)}=\int \frac{4t}{1+t}\, dt$

Rewriting the Integral: The expression $\frac{4t}{1+t}$ can be decomposed as:

$\frac{4t}{1+t}=4(\frac{{t}^{2}-1+1}{1+t})=4((t-1)+\frac{1}{t+1})$

Separate and Integrate:

$4\int (t-1)\, dt+4\int \frac{1}{t+1}\, dt$

Solving the Integrals:

The integral of $t-1$ is $\frac{(t-1{)}^{2}}{2}$.

The integral of $\frac{1}{t+1}$ is $\ln (t+1)$.

Therefore:

$f(x)=4{\frac{(t-1{)}^{2}}{2}+\ln (t+1)}+C$

Substitute $t={x}^{1/4}$: Replace $t$ back with ${x}^{1/4}$:

$f(x)=2({x}^{1/4}-1{)}^{2}+4\ln (1+{x}^{1/4})+C$

Using the Condition $f(0)=-6$:

$f(0)=2(0-1{)}^{2}+4\ln (1+0)+C=2	imes 1-0+C=-6$

Solving gives $C=-8$.

Find $f(1)$:

$f(1)=2({1}^{1/4}-1{)}^{2}+4\ln (1+{1}^{1/4})-8$

Simplifies to:

$f(1)=0+4\ln 2-8=4(\ln 2-2)$

Solutions · 1