New answers tagged indefinite-integrals
5
votes
Accepted
Compute $\int \frac{\sin(x)+\cos(x)}{e^{-x}+\sin(x)}\ dx$
Your approach is correct, you are multiplying by $e^x/e^x=1$ in the first step.
Also note that
$$
\int \frac{\sin x+\cos x}{e^{-x}+\sin x} \ dx = \int
\left( \frac{e^{-x}+\sin x}{e^{-x}+\sin x} + \...
- 21.5k
1
vote
Integral of $\int \sqrt{x+\sqrt{x^2-1}}\,\mathrm{d}x$
In fact under $u = \sqrt{x^{2} - 1} + x$ W get
$$I= \frac{1}{2} \, \int \frac{\left(u - 1\right) \left(u + 1\right)}{u^{\frac{3}{2}}} \, \mathrm{d}u =\frac{1}{2}\left[ \int \sqrt{u} \, \mathrm{d}u - \...
- 1,555
12
votes
Accepted
Integral of $\int \sqrt{x+\sqrt{x^2-1}}\,\mathrm{d}x$
Here is a solution that is very straightforward and requires NO substitutions - It seems to be the 'simple' solution that your maths teacher is looking for.
We first start with a very helpful and cool ...
- 478
0
votes
Integral of $\int \sqrt{x+\sqrt{x^2-1}}\,\mathrm{d}x$
The substitution $x=\frac12(t+t^{-1})$ mates with $\sqrt{x^2-1}$. In your case,
$$\int\sqrt{\tfrac12(t+t^{-1})+\tfrac12(t-t^{-1})}\,\tfrac12(1-t^{-2})\,dt=\tfrac12\int(\sqrt t-t^{-3/2})\,dt$$ which is ...
- 3,811
1
vote
How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?
Start off with $e^{\sqrt{x}}=u$
so $\frac{e^{\sqrt{x}}dx}{\sqrt{x}}=2du$
so you essentially have to $$2\int\sqrt{u-1}du$$
which is equivalent to $$\frac{4}{3}(e^{\sqrt{x}}-1)^{\frac{3}{2}}+C$$
- 1,491
1
vote
How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?
Let us try your approach.
$$u=\sqrt{e^{\sqrt x}-1}$$ makes $e^{\sqrt x}=u^2+1$ and $x=\log^2(u^2+1)$, then $dx=\dfrac{4u\log(u^2+1)}{u^2+1}$. So
$$\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}...
- 3,811
1
vote
How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?
A first change of variable is obvious, by
$$\int e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}\cdot2\,d\sqrt x=2\int e^u\sqrt{e^u-1}\,du.$$
Then a second change is also obvious,
$$2\int \sqrt{e^u-1}\,de^u=2\int\...
- 3,811
2
votes
Accepted
How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?
$$I=\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$$
$\sqrt{x}=t\implies \frac{1}{\sqrt{x}}\,dx=2dt$
$$I=2\int {e^{t}\sqrt{e^{t}-1}}\,dt$$
$e^t-1=z\implies e^t\,dt=dz$
$$I=2\int {\sqrt{z}}\,...
- 3,671
1
vote
How do I solve $\int \frac{e^{\sqrt{x}}\sqrt{e^{\sqrt{x}}-1}}{\sqrt{x}}dx$?
Let $e^{\sqrt x}-1=u$, so $\dfrac{e^{\sqrt x}}{2\sqrt x}\,\mathrm dx=\mathrm du$, you integral becomes:
$$I=2\int \sqrt{e^{\sqrt x}-1} \,\frac{e^{\sqrt x}}{2\sqrt x}\,\mathrm dx=2\int\sqrt u\, \mathrm ...
- 6,855
0
votes
How to integrate $\int\frac{2x+7}{4+(1+x)^2}dx$ without splitting it up?
This answer is not so different but this probably is the fastest way to solve this integral, $$I=\int\frac{2x+7}{4+(1+x)^2}dx=\int\frac{2(x+1)+5}{4+(1+x)^2}dx$$
$$I=\int\frac{2(x+1)+5}{4+(1+x)^2}dx= 2\...
- 3,671
0
votes
How to integrate $\int\frac{2x+7}{4+(1+x)^2}dx$ without splitting it up?
Notice that the derivative of $4+(1+x)^2$ is $2+2x$.
So split the integrand accordingly to
$$\int \frac{2x+2}{1+(1+x)^2}dx+ \int \frac{5}{4+(1+x)^2}dx= \ln(1+(1+x)^2)+ \frac{1}{2}\arctan(\frac{x+1}{2})...
- 1,491
2
votes
Accepted
How to integrate $\int\frac{2x+7}{4+(1+x)^2}dx$ without splitting it up?
As mentioned in the comments, one has
\begin{align*}
\int\frac{2x + 7}{4 + (x + 1)^{2}}\mathrm{d}x & = \int\frac{2(x + 1)}{4 + ( x + 1)^{2}}\mathrm{d}x + \int\frac{5}{4 + (x + 1)^{2}}\mathrm{d}x
\...
- 17.6k
2
votes
How to integrate $\int\frac{1}{x|x|+3}dx$?
For $x\ge0$,
$$\int\frac{\mathbb dx}{x|x|+3}=\int\frac{\mathbb dx}{x^2+(\sqrt3)^2}=\frac1{\sqrt3}\tan^{-1}\left(\frac{x}{\sqrt3}\right)+C$$
For $x\le0$,
$$\int\frac{\mathbb dx}{x|x|+3}=\int\frac{\...
- 637
1
vote
How to integrate $\int\frac{1}{x|x|+3}dx$?
This will be a piecewise integral in my opinion.
$$\int \frac{dx}{x^2+3} = \frac{1}{\sqrt3}\arctan\left(\frac{x}{\sqrt{3}}\right)+C_1 \quad \forall x\geq0$$
$$\int \frac{dx}{3-x^2}= \frac{1}{2\sqrt{3}}...
- 1,491
3
votes
Accepted
Simplifying the integral $\int{\frac{\ln(x)(x+1)}{(x-1)x}} \, \mathrm{d}x $
You could do partial fractions on the nonlogarithmic portion,
$$
\frac{x+1}{x(x-1)} = \frac{2}{x-1} - \frac 1 x
$$
Then you have to consider the integrals
$$
\int \frac{\ln x}{x} \, \mathrm{d}x
\qquad
...
- 47.5k
0
votes
Solving the integral of a rational function that mathematica gives a root sum for.
One can't get the the Riemann zeta function in the denominator this way because of the Euler product.
The closest one can get is feeding in logarithms of prime numbers into the vector ...
- 7,454
1
vote
Find the indefinite integral of $\int \frac{1}{\sqrt{\cos(x)(1 - \cos(x))}} \, dx$
Using the tangent half-angle substitution
$$I=\int \frac{dx}{\sqrt{\cos (x) (1 - \cos (x))}} = \int \frac{\sqrt{2}}{t \sqrt{1-t^2}}\,dt$$
$$t=\sqrt{1-u^2} \quad \implies \quad I=\int\frac{\sqrt{2}}{u^...
- 270k
1
vote
Find the indefinite integral of $\int \frac{1}{\sqrt{\cos(x)(1 - \cos(x))}} \, dx$
$$
\begin{aligned}
I & =\int \frac{1}{\sqrt{\cos x \cdot 2 \sin ^2 \frac{x}{2}}} d x \\
& =\frac{1}{2} \int \frac{1}{\sqrt{\cos x \cdot \sin \frac{x}{2}}} d x \\
& =\int \frac{d u}{\sqrt{2 ...
- 23.8k
2
votes
Accepted
Find the indefinite integral of $\int \frac{1}{\sqrt{\cos(x)(1 - \cos(x))}} \, dx$
Applying the substitution $$t = \cos x, \qquad dt = -\sin x \,dx ,$$
transforms the integral to an algebraic one: $$\int \frac{dx}{\sqrt{\cos x (1 - \cos x)}} = \int \frac{dt}{(1 - t) \sqrt{t + t^2}},$...
- 104k
0
votes
Integral $\int \frac{dx}{\tan x + \cot x + \csc x + \sec x}$
By substituting $x=y+\dfrac\pi4$ (some more details here),
$$\begin{align*}
& \int \frac{dx}{\tan x+\cot x+\csc x+\sec x} \\
&= \frac12 \int \frac{2\cos^2y-1}{\sqrt2\,\cos y+1} \, dy \\
&= ...
- 21.4k
1
vote
Evaluating $ \int\frac{\sqrt{1-x^2}-x}{\sqrt{1-x^2}(1+x\sqrt{1-x^2})}\, dx $
As an alternative to the standard tangent half-angle substitution, consider its relative
$$u=\tan\left(\frac\pi4-\frac\theta2\right) \iff \theta=\frac\pi2-2\arctan u$$
which boils down to further ...
- 21.4k
0
votes
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
$$\begin{align}(\log(s\sin x+c\cos x))'=\frac{s\cos x-c\sin x}{s\sin x+c\cos x}=\frac{s-c\tan x}{s\tan x+c}\\\\
(x)'=\frac{s\tan x+c}{s\tan x+c}\end{align}$$
From this, you can form any linear ...
- 3,811
1
vote
How to integrate $\int\frac{1}{4\tan^{2}(x)+9}dx$?
As you can verify,
$$\left(\frac pq\arctan\left(\frac pq\tan x\right)\right)'=\frac{p^2\tan^2x+p^2}{p^2\tan^2xq^2}=1+\frac{p^2-q^2}{p^2\tan^2x+q^2}.$$
Hence,
$$\int\frac{dx}{p^2\tan^2x+q^2}=\frac1{p^2-...
- 3,811
1
vote
How to integrate $\int\frac{1}{4\tan^{2}(x)+9}dx$?
In general, let $f$ be a rational function. Then the integration $$\int f(\tan x)~dx$$
can be solved by using the substitution $u=\tan{x}$. This is because $du=\sec^2x~dx=(u^2+1)dx$. Therefore, the ...
- 1,952
2
votes
Why the integral of $\sin^{-1}(x)$ doesnt exist?
By parts,
$$\int\arcsin(x)\,dx=x\arcsin(x)-\int\frac{x}{\sqrt{1-x^2}}dx=x\arcsin(x)+\sqrt{1-x^2}+C.$$
$$\int\frac{dx}{\sin x}=\int\frac{\sin x\,dx}{\sin^2 x}=-\int \frac{d\cos x}{1-\cos^2x}=-\text{...
- 3,811
4
votes
Computing the antiderivative of $\frac{1}{\sqrt{A^2+2 A B e^{C z}+B^2 e^{2Cz}-\beta^2}}$
Let’s take out all constants first and rewrite the integral as
$$
\begin{aligned}
I & =\frac{A \sqrt{(A-\beta)(A+\beta)\left(C^2+K^2\right)}}{|K|} \int \frac{d z}{\sqrt{A^2+2 A B e^{C z}+B^2 e^{...
- 23.8k
5
votes
How to integrate $\int\frac{1}{4\tan^{2}(x)+9}dx$?
$$
\begin{aligned}
\int \frac{1}{4 \tan ^2 x+9} d x
= & \int \frac{\sec ^2 x}{\sec ^2 x\left(4 \tan ^2 x+9\right)} d x \\
= & \int \frac{d t}{\left(1+t^2\right)\left(4 t^2+9\right)},\quad \...
- 23.8k
2
votes
How to integrate $\int\frac{1}{4\tan^{2}(x)+9}dx$?
Define
$$I = \int \frac{1}{4\tan(x)^2+9}dx$$
Multiply the top and bottom by $\cot(x)^2$
$$I = \int \frac{\cot(x)^2}{4+9\cot(x)^2}dx$$
Rewrite the numerator using Pythagorean identity: $\cot(x)^2=\csc(...
2
votes
How to integrate $\int\frac{1}{4\tan^{2}(x)+9}dx$?
Hint: Split the integrand as follows
$$\frac{1}{4\tan^{2}x+9}
= \frac{A(1+\tan^2x)+B(4\tan^{2}x+9)}{4\tan^{2}x+9}
= B+\frac{A\sec^2x}{4\tan^{2}x+9}
$$
and find $A$, $B$ by matching the numerator.
- 102k
0
votes
How to solve $\int\frac{1}{\sin(4x)\cos(5x)}dx$?
Employ the substitution $\sin x=\text{sech}\ t$ instead after factoring the denominator as follows, along with $dt=-\frac{dx}{\sin x}$ and $\phi=\frac{\sqrt5+1}2$
\begin{align}
&\int\frac{1}{\...
- 102k
1
vote
How to solve $\int\frac{1}{\sin(4x)\cos(5x)}dx$?
Too long for a comment. Interesting use of Chebyshev polynomial of the second kind $U_8(x)$ to convert the integrand to a rational function:
$$\int\frac{dx}{\sin4x\cos5x}=\int\frac{dx}{\sin9x-\sin x}\...
- 12.5k
1
vote
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
HINT
Here I propose another way to approach it:
\begin{align*}
\int\frac{1}{3\tan(x) - 7}\mathrm{d}x & = \int\frac{\cos(x)}{3\sin(x) - 7\cos(x)}\mathrm{d}x\\\\
& = \frac{1}{\sqrt{58}}\int\frac{...
- 17.6k
2
votes
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
Too long for a comment (you already received elegant and simple solutions)
I think that, when using the tangent half-angle substitution $x=2 \tan ^{-1}(t)$, you forgot the $dx$
$$I=\int \frac{dx}{3\...
- 270k
0
votes
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
Let $\cos x\equiv A(3\sin x-7\cos x)+B(7\sin x+3\cos x)$ for some constants $A$ and $B$, then solving
$$
\left\{\begin{aligned}
3 A+7 B & =0 \cdots (1)\\
-7 A+3 B & =1 \cdots (2)
\end{aligned}...
- 23.8k
3
votes
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
Hint. Let $$C=\int\frac{\cos x}{3\sin x-7\cos x}dx $$and $$S=\int\frac{\sin x}{3\sin x-7\cos x}dx$$
Now consider $3S-7C$ and $3C+7S$
- 34.6k
7
votes
How to integrate $\int \frac{1}{3\tan(x)-7}dx$?
In the first approach, split the integrand as
$$\frac{\cos x}{3\sin x-7\cos x}
=\frac{3}{58}\frac{3\cos x+7\sin x}{3\sin x-7\cos x} -\frac{7}{58}\frac{3\sin x-7\cos x}{3\sin x-7\cos x}
$$
- 102k
0
votes
Hint on solving $\int \frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} dx$
Split the integrand as follows
$$\frac{\sqrt{1- a^{2}+x^{2}}}{x^{2}(a^2-x^2)} = \frac{1}{a^2\sqrt{1- a^{2}+x^{2}}}\left(\frac{1-a^2}{x^2}+\frac{1}{a^2-x^2}\right)$$
Then
$$\int \frac{\sqrt{1- a^{2}+x^{...
- 102k
1
vote
Integration of $\int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx$
Result: The integral evaluates to
\begin{align}
I=&\int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx\\
=&\ \frac{r_++d}{r_+-r_-}\ln\bigg( x+\frac{\sqrt{1+x^2}}{...
- 102k
0
votes
How to integrate $\int\frac{\sqrt{16x^2-9}}{x}dx$?
$$\int \frac{\sqrt{16x^2-9}}{x}\mathrm dx$$
$$=\int \frac{16x^2-9}{x\sqrt{16x^2-9}}\mathrm dx$$
$$=16\int \frac{x}{\sqrt{16x^2-9}}\mathrm dx-9\int \frac1{x^2\sqrt{16-\frac9{x^2}}}\mathrm dx$$
For the ...
- 637
0
votes
Evaluate $\int \frac{2x}{(1-x^2)\sqrt{x^4-1}}dx$
Substitute $x^2=t$:
$$\int \frac{\mathrm dt}{(1-t)\sqrt{t^2-1}}$$
Now, put $1-t=\frac1{u}$. Note that $u=\frac1{1-x^2}<0$ as there is $\sqrt{x^4-1}=\sqrt{(x^2-1)(x^2+1)}$ in the denominator of the ...
- 637
2
votes
How to solve this integral $\int \frac 1{\sqrt { \cos x \sin^3 x }} \mathrm dx $
$$\int \frac 1{\sqrt{\cos x\sin^3x}} \mathrm dx$$
$$=\int \frac 1{\sin^2x\sqrt{\cot x}} \mathrm dx$$
Now, this can be solved easily by substituting $t=\cot x$.
- 637
0
votes
Indefinite integral: $\int \frac{\sqrt{x^2-6x+18}}{x-3}dx$
$$\int \frac{\sqrt{x^2-6x+18}}{x-3} dx$$
$$=\int \frac{(x-3)^2+9}{(x-3)\sqrt{x^2-6x+18}} dx$$
$$=\int \frac{x-3}{\sqrt{x^2-6x+18}} dx+\int \frac{9}{(x-3)\sqrt{x^2-6x+18}} dx$$
Now, substitute $x^2-6x+...
- 637
3
votes
Accepted
Integration of $\int \frac{2a^2+x^2 + b x\sqrt{a^2+x^2}}{(c+b x^2)\sqrt{a^2+x^2}+ x(a^2 +x^2)} dx$
Looking only at your final expression (i.e. making no comment on whether you took the right approach or did all the substitutions correctly), it is possible to factorise the quartic expression in the ...
- 26.2k
2
votes
Solve indefinite integral $\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}dx$
Rationalize the denominator
\begin{align*}
&\int\frac{x^2}{1-x^2+\sqrt{1-x^2}}\,dx\\
=& \int\frac{x^2}{\sqrt{1-x^2} (1+\sqrt{1-x^2} )}\,dx
=\int\frac{1-\sqrt{1-x^2} }{\sqrt{1-x^2}}\,dx
= \sin^{...
- 102k
1
vote
Best way to solve $\int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x-1}+1}$
As @Dan suggested in comments
$$u = \sqrt{x}+\sqrt{x-1}+1 \implies x=\frac{\left(u^2-2 u+2\right)^2}{4 (u-1)^2}$$ $$\implies dx=\frac{(u-2) u \left(u^2-2 u+2\right)}{2 (u-1)^3}$$
$$\int\frac{\mathrm{d}...
- 270k
2
votes
Accepted
Lengthy Indefinite Integral $\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx$
Split the integrand into two
and, then, apply to both the 3rd Euler substitution $ t=\frac{\sqrt{(x-3)(x-1)}}{x-1} $
\begin{align}
&\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx\\
=& \int \frac1{\...
- 102k
0
votes
Evaluating the integral $\int \frac{x-1}{(x+1) \sqrt{x^3+x^2+x}}dx$
$$
\begin{aligned}
I& =2 \int \frac{y^2-1}{\left(y^2+1\right) \sqrt{y^2+y^2+1}} d y, \quad \text { where } y=\sqrt{x} \\
& =2 \int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right) \sqrt{y^2+\...
- 23.8k
2
votes
Lengthy Indefinite Integral $\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1} dx$
$$I=\int \frac{\sqrt{x^2-4x+3}}{x^2+x+1}\, dx$$
Using Euler substitution
$$\sqrt{x^2-4x+3}=t+x \implies x=\frac{3-t^2}{2 (t+2)}\implies dx=-\frac{t^2+4 t+3}{2 (t+2)^2}$$
$$I=-\int \frac{(t+1)^2 (t+3)^...
- 270k
1
vote
Integral of $\int \frac{\sqrt{x^2-1}}{x}dx$
$$\int \frac{\sqrt{x^2-1}}{x} dx$$
$$=\int \frac{x^2-1}{x\sqrt{x^2-1}} dx$$
$$=\int \frac{x}{\sqrt{x^2-1}} -\frac{1}{x\sqrt{x^2-1}}dx$$
$$=\sqrt{x^2-1}-\sec^{-1}x+C$$
- 637
4
votes
Best way to solve $\int\frac{\mathrm{d}x}{\sqrt{x}+\sqrt{x-1}+1}$
Let $t^2=x-1$, then
$$
\begin{aligned}\int \frac{d x}{\sqrt{x}+\sqrt{x-1}+1}
= & \int \frac{2 t d t}{\sqrt{t^2+1}+t+1} \\
= & \int \frac{2 t\left(t+1-\sqrt{t^2+1}\right)}{(t+1)^2-\left(t^2+1\...
- 23.8k
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